Solutions of Selected Examples for Practice...Solutions of Selected Examples for Practice Example...

Solutions of Selected Examples for Practice

Example 0.6.2

Solution : Kept this unsolved example for student's practice.

Example 0.7.2


Example 0.7.3

Solution : The cylinder is shown in the Fig. 1.

Consider a differential volume in the cylindrical co-ordinate

system as,

dv = r dr d dz�For the given cylinder r varies from 0 to R, z varies from 0 to

L while � varies from 0 to 2 � radians. These are the limits of

integration.

� Volume of cylinder =

0 0

2

0

L R

r dr d dz� � � ��

=

0

L Rr

d dz� ��

��

��

�0

2

02

��

= �R

dzL2

002

2 � � �

= �22

2� �

Rz 0

L � R L2

Example 0.7.4

Solution : Consider the upper surface area, the normal to which is a z . So the differential

surface area normal to z direction is r d dr� . Consider the Fig. 1.

� S1 =

0 0

��

� � �R

r d dr =

0

��

��

��

��

�r2

d2

0

R

= �R2

02

2� � �� R2

The bottom surface area S2 is same as S1 i.e. �R2 . For remaining surface area consider the

differential surface area normal to r direction which is r d dz� .

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM

(0 - 1)

0 Vector Analysis

x

y

z

ar

L

az

R

S3

S1

S2

Fig. 1

S 3 =

0 0

L

r d dz� � ��

but r R� is constant = � �0 0

02

L

0LR d dz R z� � � � � �

�� 2 RL

Total surface area = S S S R R RL1 2 3� � � � �� 2 2 2 = � �2 R R L� �

Example 0.8.2

Solution : Consider the differential surface area normal to the r direction which is,

dSr = r d d2 sin � � �

Now the limits of � are 0 to 2 � while � varies from 0 to � .

� Sr =

0 0

��

� �� r d d2 sin

But note that radius of sphere is constant, given as r R� .

Sr = � �R d d R2 2

0 00 0

2��

� �� sin cos

= � � � � � � � �R R22 0 2 1 1 2� � � � � � � �cos cos�� = 4 R2�

Example 0.8.3

Solution : Consider the spherical shell of radius a hence r a� is constant.

Consider differential surface area normal to r direction which is radially outward.

dSr = r d d a d d2 2sin sin� � � �� ... as r a�

But � is varying between 0 to � while for spherical shell � varies from 0 to �.

� Sr = a d d a2

0 0

20 0

� �� sin [ cos ] [ ]

= � �a 2 � � �[ cos cos ]�� 0 2 a 2

So area of the region is 2 a 2�.

If � � � �, the area of the region becomes 4 �a 2 , as the shell becomes complete sphere of

radius a when � varies from 0 to 2 � .

Example 0.10.2

Solution : The dot product is,

A B� = A B A B A Bx x y y z z� � = � � � �� 2 3 5 5 4 2� � � � � = 6 25 8� �

= �27As A B� is negative, it is expected that the angle between the two is greater than 90�.

|A| = � � � � � �2 5 42 2 2� � � � = 45, |B| = � � � � � �3 5 22 2 2� � = 38

� � = cos� ��

��

1 A B|A| |B|

= cos��

��

��

1 27

45 38= 130.762�

Electromagnetic Theory and Transmission Lines 0 - 2 Vector Analysis


Example 0.11.2

Solution : The perpendicular vector to the plane containing A and B is given by their

cross product.

A B� =

a a ar z�

�

�

A A A

B Br z

r z�

=

a a ar z�

��

2 1

� �1 2 �

= � � �72

3 4�

��a a ar z

� a n = Unit vector in the direction A B�

=� � �

!"#

$%& � �

72

3 4

72

3 42

2 2

��

��

�a a ar z

( ) ( )

=� � �3.5

16.9651

� ��a a ar z3 4= � � �0.648 a 0.1768 a 0.74 ar z�

Example 0.13.4


Example 0.13.5


Example 0.13.6

Solution : From the given field W,

Wx = 10, Wy = – 8 and Wz = 6

Now Wr = �W a a a a ar x y z r� �� 10 8 6

= 10 8 6a a a a a ax r y r z r� � ��

= � � � � � �10 8 6 0cos sin� � � � ... Refer Table 0.13.1

For point P, x = 10 and y = – 8

� � = tan �1 y

x... Relation between cartesian and cylindrical

= tan � ��

�� 1 8

1038.6598

As y is negative and x is positive, � is in fourth quadrant. Hence � calculated is correct.

� cos � = 0.7808 and sin � = – 0.6246

� Wr = � � � �10 0.7808 8 0.6246 12.804� � � � �

Now W� = W a a a a a a ax y z� � � �� 10 8 6

= � �10 8 0� � � ��sin cos = 0

And Wz = W a a a a a a az x z y z z z� � � �� 10 8 6

= 10 0 8 0 6 1 6� � � � � �

� W = 12.804 a ar z� 6 in cylindrical system.



Example 0.13.7

Solution : The given vector is in cylindrical system.

� Hx = H a a a a a a ax r x x z x� � � �� 20 10 3

= � �20 10 0cos sin�� ... Refer Table 0.13.1

At point P, x = 5, y = 2 and z = – 1

Now � = tan tan� �� 1 1 25

y

x21.8014

� cos � = 0.9284 and sin � = 0.3714

� Hx = � �20 0.9284 10 0.3714 22.282� � � �

Then Hy = H a a a a a a ay r y y z y� � � �� 20 10 3

= 20 10 0sin cos��

= � � � �20 0.3714 10 0.9284 1.856� � � � �

And Hz = H a a a a a a az r z z z z� � � �� 20 10 3

= 20 0 10 0 3 1 3� � � � � �

� H = 22.282 a x � 1.856 a y + 3 a z in cartesian system.

��



Solutions of Selected Examples for Practice

Example 1.2.9

Solution :

R = [(– 3) – (+ 3)] a x + [(1) – (8)] a y + [(4) – (– 2)] a z

= – 6 a x – 7 a y + 6 a z , |R| = 36 49 36� � = 11

� a R =� � �6 7 6

11

a a ax y z

� F2 =Q Q

4 R

1 22��0

a R

=33 10 10 10

4 11

6 6

02

� � �

�

� �

�� ( )

� � ��

�

�

6 7 6

11

a a ax y z

� = – 0.0133 a x – 0.0155 a y + 0.0133 a z

Example 1.2.12

Solution. : The arrangement of charges is

shown in the Fig. 1.2. The charge at P is test

charge i.e. QP = 1C.

F1P =Q Q

R

P1

0 124��

a R1 ... Force due to Q1

R1 = � � � �� 1 1 1 2 0 0a a ax y z

= – 2 a x – a y

� R1 = R1 � � �2 1 52 2

� F1P =

� �Q 1

02

4 5

2

��

� ��

�

�

a a

5x y


(1 - 1)

1Coulomb’s Law and

Electric Field Intensity

P1

Q1Q2

P (–3,1,4)2

(3,8,–2)

R

aR

Fig. 1.1

z

y

x

0

P (–1, 1, 0)

Q (1, 2, 0)1Q (2, 0, 0)2 R2

R1

Fig. 1.2

= � �Q 1

04 5 52

�� a ax y

F2P =Q Q

R

P2

0 224��

a R2 ... force due to Q2

R2 = [(–1) – (2)] a x + [1 – 0] a y + [0 – 0] a z = –3 a x + a y

R2 = R2 = � � � �3 1 102 2� �

� F2P =

� �� Q Q2

02

2

04 10

3

10 4 10 103

��

� ��

�

� � � �

a aa a

x yx y

� FP = F1P + F2P

=1

4

2

5 5

3

10 10 5 5 10 100

1 2 1 2

��

��

��

�

��

��

�

��

�

��

� Q Q Q Qa ax y�

�

��

i) If x component of FP must be zero then,

��

2

5 5

3

10 101 2Q Q

= 0 i.e. Q1 = – 0.5303 Q2

ii) If y component of FP must be zero then,

��

Q Q1 2

5 5 10 10= 0 i.e. Q1 = 0.3535 Q2

Example 1.2.13

Solution : The charges are shown in the Fig. 1.3.

F21 =Q Q

4 R

1 2

0 AB2��

aAB

RAB = (2 0) (3 0) (6 0)� � � � �a a ax y z= � � �2 3 6a a ax y z

|R |AB = 4 9 36� � = 7

aAB =R

|R |

a a aAB

AB

x y z2 3 6

7�

� � �

� F12 =0.7 10 4.9 10

4 8.854 10 7

3 6

12

� � �

� � �

� �

�� 2

2 3 6

7

a a ax y z� ��

�

�

= 0.1797 a 0.2696 a 0.5392 a Nx y z� �

|F |12 = (0.1797) (0.2696) (0.5392)2 2 2� � = 0.6291 N … Magnitude

Electromagnetic Theory and Transmission Lines 1 - 2 Coulomb's Law and Electric Field Intensity


RAB

Q = 4.9 C2 �

Q = 0.7 mC1

A (0, 0, 0)

B (2, 3, 6)

x

y

z

Fig. 1.3

Example 1.2.14

Solution : The charges are shown in the Fig. 1.4.

The position vectors of the points A, B and C

are,

A = � �4 3a ay z

B = a ay z�

C = � �3 4a ay z

� R1Q = C A a ay z� � �

and R2Q = C B a ay z� � � �4 3

� R1Q = � � � �1 1 22 2� � and

R2Q = � � � �� 4 32 2 = 5

� F1 = Force on Q due to QQ Q

R1

1

1Q2

�4 ��

a 1Q

and F2 = Force on Q due to QQQ

R2

2

2Q2

�4 ��

a 2Q

� Ft = F F a a1 2 1Q 2Q� � ��

�

�

Q Q

R

Q

R

1

1Q2

2

2Q24 ��

=

� � � �

Q Q

5

224

2 10

2 2

4 3

5

9

2��

� ��

��

�

��

� ��

��

�

��

� � a a a ay z y z

�

�

= � � � �Q7.071 10

Q

12510 2

44 3

��

�

�

� � a a a ay z y z

� Total z component of Ft is,

=Q

7.071 103 Q

12510 2

4 ��

�

�

� � a z

To have this component zero,

7.071 103 Q

12510 2� �� = 0 as Q is test charge and cannot be zero.

� Q 2 = �� 7.071 10 125

3

10= – 29.462 nC



x

O 1–3–4

A

C

B

Q1

Q2

Q

R1Q

R2Q

y

z

Fig. 1.4

Exmaple 1.2.15 :

Solution : The charges are shown in the

Fig. 1.5.

Consider the charge at P (0, 0, 2 2)4 i.e.

P (0, 0, 2.828)4 .

Let us find force on charge at P4 due to all

the charges at P , P1 2 and P3 .

1. Force due to charge at P1.

F1 =Q Q

4 R

1 4

0 12��

a R1 =Q Q

4 R

1 4

0 12��

R|R |

1

1

�

�

� … Q Q 1 mC1 4� �

R1 = (0 2) (0 0) (2.828 0) 2� � � � � � �a a a ax y z x 2.828� a z

� F1 =1 10 1 10

4 8.854 10 [2 2.

3 3

12 2

� � �

� � � �

� �

�� 828 ]

2 2.828

2 2.8282 2 2

� �

�

�

�

�

a ax z

= � �432.423 611.446 Na ax z

2. Force due to charge at P2 .

F2 =Q Q

4 R

Q Q

4 R

2 4

0 22

2 4

0 22��

aR

|R |R22

2�

�

�

�

R2 = [(0) ( 1)] [0 3] [2 2 0]� � � � � � � �a a a ax y z x 1.732 2.828a ay z�

� |R |2 = 1 1.732 2.828 12 3.46412 2 2� � � �

� F2 =1 10 1 10

4 8.854 10 12

3 3

12

� � �

� � �

��

��

a x 1.732 2.828

12

a ay z��

�

�

= 216.211 374.489 611.446 Na a ax y z� �

3. Force due to charge at P3

F3 =Q Q

4 R

Q Q

4 R

3 4

0 32

3 4

0 32��

aR

| R |R33

3�

�

�

�

R3 = [0 ( 1)] [0 ( 3)] [2 2 0]� � � � � � � �a a a ax y z x 3 2 2� �a ay z

� |R |3 = 1 ( 3) (2 2) 12 3.46412 2 2� � � �

� F3 =1 10 1 10

4 8.854 10 12

3 3

12

� � �

� � �

��

��

a x 3 2 2

12

a ay z��

�

�

= 216.211 374.489 611.538 Na a ax y z� �



P (0, 0, 2 2 )4

P (–1, 3, 0)2P (–1, – 3, 0)3

P (2, 0, 0)1

z

x

y

R3

R2

R1

Fig. 1.5

� The total force on charge at P4 is,

F = F F F a a1 2 3 z z3 611.446 1834.338 N� � � � � i.e. |F| = 1834.338 N

The magnitude of force on each charge remains same as above due to symmetrical

distribution of charges.

Example 1.2.16

Solution : The arrangement is shown in the

Fig. 1.6.

The co-ordinates of the vertices of triangle are,

Point O ! (0, 0, 0)

Point Q ! (0.1, 0, 0)

Point P ! (0.05, 0.0866, 0).

Let us find the force on P due to the charges at

O and Q.

� F1 =Q Q

4 R

Q Q

4 R |1 2

0 OP2

1 2

0 OP2��

aR

ROPOP

OP� �

|

� ROP = 0.05 0.0866a ax y� , | | (0.05) (0.0866) 0.12 2ROP � � �

� F1 =0.25 10 0.25 10

4 8.854 10

6 6

12

� � �

� � �

� �

�� 0.1

[0.05 0.0866 ]

0.12�

�a ax y= 0.02808 0.0486 Na ax y�

F2 =Q Q

4 R

Q Q

4 R | |1 2

0 QP2

1 2

0 QP2��

aR

RQPQP

QP� �

RQP = (0.05 0.1) 0.0866 0.05 0.0866� � � � �a a ax y x a y

|RQP|= (0.05) (0.0866)2 2� = 0.1

� F2 =0.25 10 0.25 10

4 8.854 10 0

6 6

12

� � �

� � �

� �

�� .1

[ 0.05 0.0866

0.12�

� �a a ]x y= � �0.02808 0.0486 Na ax y

� F = F F 0.09729 a N1 2 y� � … Direction a y

� |F| = 0.09729 N … Magnitude

Example 1.3.10

Solution : Consider the circle consisting of charges placed in xy plane and charge of

�20 �C is on z axis, 2 m from the plane of the circle. This is shown in the Fig. 1.7.

The charges are placed equally i.e. at an interval of 360 10"/ = 36" between each other. Five

pairs of charges which are dimetrically opposite to each other, exists on the circumference



y

xO

d

10 cm 10 – 52 2

= 8.66 cm

Q

5 cm

10 cm

5 cm 0.25 C�0.25 C�

0.25 C�P

R

Fig. 1.6

of a circle. Consider a pair A and B. The

field EA due to Q at A, at point P is shown

in the Fig. 1.7.

l (OQ) = 2 m, l (OP) = 2 m

hence # PAO = 45"

� y component of EA i.e. E EA y A� "cos 45

Similarly l (OB) = 2 m, l (OP) = 2 m

hence # PBO = 45"

� y component of EB i.e. E EBy B� "cos 45

But EAy is in �a y direction while EBy is in

a y direction. From symmetry of the

arrangement E EAy By� . Hence they cancel

each other.

While z components of EA and EB help each other as both are in a z direction.

EAz = � �E E E aBz A B z� "or sin 45

Similarly there are 4 more pairs of charges which will behave identically and their y

components are going to cancel while z components are going to add.

Thus total z component of E at P is,

Etotal = (E due to any charge) � � "10 45sin a z =Q

R2410 45

�� "sin a z

where R = � � � �2 2 82 2� �

� Etotal =

� �500 10

4 810 45

6

2

�

�� "

�

��

sin a z = 3.972 � 106 a z V/m

� FP = Q 10 3.972 10P6 6E atotal z� � � � ��20 = – 79.44 ( )a z N

This is the force on the charge at P. In general, force acts normal to the plane in which

circle is kept, i.e. – 79.44 a n where a n is unit vector normal to the plane containing the

circle.

Example 1.3.11 :

Solution : a) A (2, –1, 3) and P (0, 0, 0)

� E at P =Q

RAP24 ��

aAP



Oy

x

z

45º 45º

2 2

2

ABQ

Q

QQ Q

Q

Q QQ

Q

R R

P(0,0,2)EAy

EA

EBy

EB

45º45º

Fig. 1.7

Now aAP =� � � ��

� � � � � �

0 2 0 1 0 3

2 1 32 2 2

� � � � � �

� � �

a a ax y z

� E =

� �5 10

4

2 3

14

9�

� � �

� � ��

�

�

�

�� 8.854 10 1412 2

a a ax y z

= – 1.715 a x + 0.857 a y – 2.573 a z V/m

b) Let point P is now (x, 0, 0).

� aAP =� �

� � � � � �

rr

a a aAP

AP

x y z�

� � �

� � � �

x 2

x 2 1 32 2 2

3

� E =� ��

� �

� �

Q

x 2 1 9

x 2

x 2 1 92 24

3

�� $

� � �

� � �

�

�

�

a a ax y z

=

� �� 5 10

x 2 10

x 29

2 3/ 2

�

� ��

�

4

3

��

a a ax y z

=

� �� 44.938

x 2 10

x 22 3/ 2

� �� a a ax y z3

| |E =

� ��

44.938

x 2 10

x 2 1 32 3/ 2

2 2 2

� ��

��

=� ��

44.938

x 2 102� �V/m

To find x at which| |E is maximum,

d

dx

| |E= 0

� 44.938� �

� ��

� �

�

�

�

2 x 2

x 2 102 2= 0 i.e. (x – 2) = 0

� x = 2 where| |E is maximum.

The graph of| |E against x is shown in the

Fig. 1.8.

c) Hence| |maxE is at x = 2,

� | |maxE =44.938

10= 4.4938 V/m



4.49| |E max

| |E in V/m

10–10 20x

Fig. 1.8

Example 1.3.12 :

Solution : The charges Q 1 and Q 2 are shown in the Fig. 1.9.

Let us find E at the origin due to Q 1 and Q 2 .

P1 = �4 a y , P a2 z� 4

� a 1 =� ��

� ��

�P

|P

a

4a1

1

yy

|

4

and a 2 =�

��

� �P

|P

a4

a2

2

zz

|

4

RP 1O = 4 and RP2O = 4

� E = E E a a1 2 1 2� � �Q

R

Q

R

1

P1 O2

2

P2 O24 4��

= � ��

� �10 10

4 4

20 10

4 4

9

2

9

2

�

��

�

��

� �

��

a ay z = 5.6173 a y – 11.2346 a z V/m

Now let Q 3 = 40 nC is at point � �P x, y, z3 .

� P3 = x a x + y a y + z a z and RP3O = x y z2 2 2� �

The field intensity due to Q 3 at the origin is,

E3 =Q

R

3

P3O2 O

4 �� a P3

=Q

R

x y z

x y z

3

P3O2 2 2 24 ��

� � �

� �

�

�

�

a a ax y z

The total E has to be zero with E3 added to E1 and E2 .

� E E E1 2 3� � = 0

In E E1 2� , there is no x component and to have x component of E with E3 zero, x = 0.

The y component of E3 must cancel y component of E E1 2� .

� ��

y Q

R x y z

3

P3O2 2 2 24 ��

= – 5.6173

Now RP3O = x y z2 2 2� �

��

y Q

x y z x y z

3

2 2 2 2 2 24 �� = 5.6173



y

x

a1

a2

z

OP

(0,–4,0)1

P (0,0,4)2Q2

Q1

Fig. 1.9

�

� �y

x y z2 2 2 3/ 2� �

=5.6173�

� �

4

40 10 9

�� = 0.01562 ... Q 3 = 40 nC

But x = 0 hence,

� �y

y z2 2 3/ 2�

= 0.01562 ... (a)

Similarly z component of E3 must cancel z component of E E1 2� .

��

� �

z Q

R x y z

3

P3O2 2 2 24 ��

= 11.2346

Substituting Q , R3 P3O and x = 0 we get,

�

� �z

y z2 2 3/ 2�

= – 0.03124 ... (b)

From equation (a), � �y z2 2 3/ 2� =

y

0.01562= 64.0204 y

Putting in equation (b),z

64.0204 y= – 0.03124 i.e. z = – 2 y ... (c)

Using equation (c) in (a),

� �� y

y 2y2 2 3/ 2� �

= 0.01562 i.e.

� �y

y 4 y2 2 3/ 2�

= 0.01562

Solving, y = % 2.3929 and z = � 4.7858

But y must be positive and z must be negative in P3 to have E = 0.

�Q3 must be located at (0, 2.3929, – 4.785)

Example 1.3.13

Solution : E =Q

R024��

a R

a R =R

RQP

QP=

P Q

P Q

�

�

P Q� = � � � � � �� 0 2 0 2 0 01 2 3 2 5. . . . .a a ax y z

= � � �0 4 01 0 2. . .a a ax y z

� a R =

� � � � � �

� � �

� � �

0 4 01 0 2

0 4 01 0 22 2 2

. . .

. . .

a a ax y z



y

z

xaR

P

Q5 nC

Fig. 1.10

=� � �0 4 01 0 2

0 45825

. . .

.

a a ax y z

= � � �0 8728 0 2182 0 4364. . .a a ax y z

� R = P Q� = 0.45825

� E =� �

5 10

4 8 854 10 0 45825

9

12 2

�

� � �

�

�� . .� �a R = 214 a R

Substituting value of a R ,

E = � � �186.779 a 46.694 a 93.389 ax y z V/m … E at P

Example 1.3.14

Solution : The arrangement is shown in the Fig. 1.11. Let

the charges are placed at A, B and C while E is to be

obtained at fourth corner O.

� E at 'O' = E E EA B C� �

=Q

4 R

Q

4 R

Q

4 R0 A2

0 B2

0 C2��

a a aRA RB RC� �

RA = – 0.05 a x , RA = 0.05, a RA = – a x

RB = – 0.05 – 0.05a ax y , RB = 0.0707,

a RB = – 0.707 – 0.707a ax y

RC = – 0.05 a y , RC = 0.05, a RC = – a y

� E =Q

4

–

(0.05)

– 0.707 – 0.707

0.07070 2��( ) ( )

( )

a a ax x y�

2�

�

�

�

( )–

(0.05) 2

a y

=100 10

4 8 854 10400 141 4 141 4 40

9

12

�

� �

–

–.– – . – . –

�a a ax x y� �0 a y

= – . – .486 6 486 6a ax y kV/m, |E| at 'O' = 688.156 kV/m

Example 1.3.15

Solution : The arrangement is shown in the Fig. 1.12.

F3 = F F13 23� =Q Q

4 R

Q Q

4 R

1 3

0 132

2 3

0 232��

a aR13 R23�

R13 = – 3 2a a ax y z� � , R13 = 14, a R13 =R

R13

13



x

z

y

AB

CO

RA

RC

RB

(0,0.05,0)(0,0,0)

(0.05,0,0) (0.05,0.05,0)

Fig. 1.11

x

z

y

R13

(–1, ,4)–1

(0,3,1)

R23

Q = – 2 mC2

Q = 1 mC1

Q = 10 nC3

(3,2, )–1

Fig. 1.12

R23 = a a ax y z� 4 3– , R23 = 26, a R23 =R

R23

23

� Fa a a

3x y z

��

��

�

��1

41 10 10 10

14 14

3 9

2��

– –

( )

– 3 + + 2� �

� � � �

��

�

��

�

�

�

–

( )

– –2 10 10 10

26 26

3 9

2

a a ax y z+ 4 – 3

= – 6.503 a 3.707 a 7.4985 ax y z– � mN

E3 =FQ

3

3= – 650.3 a 370.7 a 749.85 ax y z– � kV/m

Example 1.4.3

Solution : i) 0 < x < 5 m, &L = 12x mC m2

Q = &L dL' = 12 x dx mC2

0

5

' = 12x3

3

0

5�

�

�

= 500 mC = 0.5 C

ii) &S = &z nC m2 2 , & = 3, 0 < z < 4 m

Q = &SS

dS' = � �& & (SS

d dz' = & & ((

�

z

2 –9z 10 d dz

� �' ' �

0

4

0

2

... & = 3

= � � � �3 10z3

2 –902

3

0

4

� ��

�

�

( � = 1.206 C�

iii) &v =10

rsinC m 3

), r = 4 m

Q =

vol

v dv' & =

vol

v2r sin dr d d' & ) ) ( =

(

�

)

� *

)) ) (

� � �' ' ' �

0

2

0 0r

210r sin

r sin dr d d

= � � � �102

2

0

4

0 02r�

�

�

) (� � = 1579.136 C.

Example 1.4.4 :

Solution : n e =1000

rcos

4(

electrons/m 3

1 electron = � � �1.6 10 19 C charge

� &v = n e � charge on 1 electron =� � �1.6 10

rcos

4C/ m

163(

The volume is defined as sphere of r = 2 m.

� dv = r sin dr d d2 ) ) ( ... spherical system



� Q =

volv

r 0

2 162dv

1.6 10r

r' ' ' '��

� � �

�&

(

(

�

)

�

0

2

04

cos sin dr d d) ) (

= � ��

�

�

�

�

�1.6 10r2

cossin

414

162

0

2

0)

(�

�

�

0

2 �

= � � � � � � � � �� 1.6 10 2 2 4 1 2.56 1016 15 C

Example 1.4.5

Solution : &v = 0.2 �C m 3 , dv = r sin dr d d2 ) ) ( for spherical

i) Q =

v' ' ' '� �

� � �

&(

�

)

�

v

0

2

0 r 0.03

0.05

dv (0.2 10�6 2) r sin dr d d) ) (

= [0.2 10 ]r3

[ cos ] [ ]63

0.03

0.05

0 02�

�

�

�

�� ) (� �

= [0.2 10 ]r3

[ cos cos ] [63

0.03

0.05

��

�

�

� �� 0 2 ] = 82.0978 pC

ii) To find r1 where Q =12

82.0978 41.049 pC� �

� 41.049 10 12� � = (0.2 10 )r3

[ cos cos 0] [263

0.03

r 1

��

�

�

� �� ]

� 4.8998 10 5� � = r (0.03) i. e. r 7.5998 1013 3

13 5� � � �

� r1 = 0.0423 m = 4.235 cm

Example 1.6.7

Solution : i) For origin let r = r1

E =&��

L

r2 0 1a r1

Point on the line is (x, 3, 5). Origin is (0, 0, 0)

Do not consider x co-ordinate as the charge is parallel to x-axis.

� r1 = (0 – 3) a y + (0 – 5) a z

= – 3 a y – 5 a z , |r 1| = 34

� E =30 10

2 34

3 5

34

9�

� � �

� ��

�

�

�

� 8.854 10-12

a ay z



= – 47.582 a y – 79.303 a z V/m

ii) P(5, 6, 1)

� r2 = (6 – 3) a y + (1 – 5) a z = 3 a y – 4 a z , |r2 |

= 5

� E =30 10

2 8.854 10 5

3 4

5

9

12

�

� � �

��

�

�

�

��

a ay z

= 64.711 a y – 86.2823 a z V/m

Example 1.6.8

Solutino : a) The line charge is shown in the

Fig. 1.14.

It is parallel to the x axis as y = 1 constant

and z = 2 constant. The line charge is infinite

hence using the standard result,

E =&

�� L

2 ra r

To find a r , consider a point on the line charge

(x, 1, 2) while P (6, –1, 3). As the line charge

is parallel to x axis, do not consider x

coordinate while finding a r .

� r = � � � �� 1 1 3 2 2a a a ay z y z , r = � � � �� 2 1 52 2

� a r =r

|r|

a ay z�� 2

5

� E =� �&

��

L122 8.854 105

2

5

24 10 2

2

9� ��

�

� �

� � �

� �

�

�

a a a ay z y z

�5= – 172.564 a y + 86.282 a z V/m

b) Consider a point charge QA at A (–3, 4, 1).

The electric field due to QA at P (6, –1, 3) is, EA =Q

R

A

AP24 ��

aAP

RAP = � �� 6 3 1 4 3 1 9 5 2� � � � � � � � � �a a a a a ax y z x y z , RAP = 10.488

� aAP =R

R

a a aAP

AP

x y z�

� �9 5 2

10.4888



z = 5

z

Oy = 3

y

x

Parallelto x-axis

Fig. 1.13

rP(6,–1,3)

1

2–+

+ O y

x

z

Fig. 1.14

� EA =� �

Q

10.4888 10.4888A

24

9 5 2

��

� ��

�

�

a a ax y z

The total field at P is now, Et = E+ EA

The y component of total Et is to be made zero.

��

� ��

�

�

�

172.564Q

10.4888

A3

5

4 �� a y = 0 i.e.

� �

5

4

Q

10.4888

A3��

= – 172.564

� QA =� �� 172.564 8.854 10 10.488812 34

5

�= – 4.4311 �C

Example 1.6.9

Solution : The charge is shown in the Fig. 1.15

Key Point Charge is not infinite hence basic method of

differential charge dQ must be used.

� dQ = &L dL = &L dz

� dE =dQ

4 R2��0

a R =&

��L dz

4 R20

R|R|

i) To find E at origin

� R = – z a z , |R| = z, a R = – a z

� dE =&

��L

o

dz( )

4 z2

�a z i.e. E =�

�'

&��

L

o4dz

z2z 1

3

a z

� E =� �

� ��

��

�

� �

�20 10

4

19

1

3

� 8.854 10 12 z z

z

a z = – 119.824 a z V/m

ii) To find E at P(4, 0, 0)

� R = (4 – 0) a x + (0 – z) a z = 4 a x – z a z , |R| = 16 2� z

Key Point As the charge is not infinite, all the co-ordinates are considered.

� dE =&

��L

o

dz

4 (16+ z2 )

4 z

16 z2

a ax z�

�

�

�

�

� E =&��

L

o44 dz

16 z

dz

16 z2 2

a ax z

( ) ( )/ /��

�

�

�

�

�� ' '3 2

1

3

3 21

3

z z

z



z

z = 3

dQ (0,0,z)

z = 1

y

x

Fig. 1.15

I1 I2

For I1, use z = 4 tan ), dz = 4 sec2 ) d)

Limits : z = 1, )1 = 14.03º and z = 3, )2 = 36.87º

I1 =4 4

16 16

2

2 3 21

2 �

�'

sec

( tan ) /

) )

))

)d

=4

4

2 2

3 31

2 sec

sec

)

))

)

' d) =14

1

2

cos )

)

)

' d)

=14 1

2[sin ]) ))

=14

[sin 36.87º – sin 14.03º] = 0.08938

Using same substitution for I2 we get,

I2 = –4 4

4

2

31

2 tan ) ) )

))

)�

'sec

sec 3

d= � � �'

sincos sec

)) )

)

)1 1

41

2

d)

= � 'sin )

)

)

41

2

d) = –14 1

2[ cos ]� ) ))

= –14

[– cos 36.87º + cos 14.03º]

= – 0.04253

� E =20 10

4 10

9

12

�

� �

�

�� 8.854[ 0.08938 a x – 0.04253 a z ] = 16.066 a x – 7.645 a z V/m

Example 1.6.10

Solution : i) Charge is infinite along z-axis.

� E =&��

L

or2a r , Point on line charge (0, 0, z), P (1, 2, 3)

� r = (1 – 0) a x + (2 – 0) a y = a x + 2 a y , |r| = 5

Do not consider z co-ordinate as charge is along z axis and infinite.

� E =2 10

2 5

2

5

6�

�

��

�

�

�

��o

a ax y= 7.19 a x + 14.38 a y kV/m

ii) Charge is finite from z = – 4 to z = + 4

Refer Ex. 1.6.9 for the procedure.

dE =&��L

o

dz

4

a a ax y z� � �

� �

�

�

�

2 3

5 3 2 3 2

( )

[ ( ) ] /

z

z

Integrating for z = – 4 to + 4 and using the substitution 3 – z = 5 tan ), the final answer

is,

E = 4.891 a x + 9.782 a y + 4.891 a z kV/m



Example 1.6.11

Solution : The line is shown in the Fig. 1.16.

The line with x = – 3 constant and y = 4

constant is a line parallel to z axis as z can take

any value. The E at P (2, 3, 15) is to be

calculated.

The charge is infinite line charge hence E can be

obtained by standard result,

E =&��

L

2 ra r

To find r, consider two points, one on the line

which is (–3, 4, z) while P (2, 3, 15). But as lineis parallel to z axis, E cannot have componentin a z direction hence z need not be consideredwhile calculating r.

� r = � �� 2 3 3 4 5� � � � � �a a a ax y x y ... z not considered

� | |r = � � � �5 1 262 2� � �

� a r =rr

a ax y

| |�

�5

26

� E =� �&

��

L

2 8.854 10$

��

�

� �

� �

� �

�

�

1

26

5 25 10 5

2

9a a

26

a ax y x y

12 26�

= 86.42 a x � 17.284 a y V/m

Example : 1.6.12

Solution : The charge is shown in the Fig. 1.17.

Key Point As charge is along z-axis, E can not have any

component in az direction.

Do not consider z co-ordinate while calculating r .

� r = ( 2 0) (2 0)� � � �a ax y

= � � � � �2 2 , 4 4 8a a rx y

� E =&� �

&� �

L

0

L

02 r 2 ra

rrr �



–34

O

P(2,3,15)

– +

+z

y

x

ar

r

Fig. 1.16

z&L = 40 nC/m

, P(–2,2,8)

x

(0,0,z)

Fig. 1.17

=� �40 10 2 a 2 a

2 8.854 10 8 8

9x y

12

� � �

� � � ��

�

��179.754 a x �179.754 a V my

Example 1.6.13

Sol. : Consider the charge along z-axis as shown in

the Fig. 1.18 Consider the differential charge at a

distance z.

dQ = &L dl = &L dz

� dE =&

��L

02

dz

4 Ra R

R = 0 ( h 0) (0 z)a a ax y z� � � � �

= � �h za ay z

| |R = h z| |

2 2R� �, a

RR

� dE =&

��L

02 2 2 2

dz

4 (h z )

h z

h z�

� �

�

a ay z

� E =&

��L

0 z1

z2

2 2 3/ 2z1

z2

2 24

h dz

(h z )

z dz

(h z' '

�

��

�

a ay z

) 3/ 2

I1

�

�

�

-I2

-

…(1)

I 1 =

z1

z2

2 2 3/ 2

h dz

(h z )'

�, z = h tan ), dz = h sec2 ) d)

I 1 =

z1

z2 2 2

3

h sec d

h'

) )

)sec 3=

1h

z1

z2

' cos ) )d =1h z1

z2[sin ])

=1h

z

h z2 2z1

z2

�

�

�

�

=1h

z

h z

z

h z

2

222

1

212�

��

�

�

�

I2 =

z1

z2

2 2 3/ 2

z dz

(h z )'

�, h2 + z2 = u2, 2z dz = 2u du

I2 =

z1

z2

3

u du

u' = ��

��

1u z1

z2= �

�

�

�

�

1

h z2 2z1

z2



P(0, –h,0)

(0, 0, z )2

(0, 0, z )1

R

y

A

B

z

dl

x

z

(0, 0, z)

Fig. 1.18

)h

z.h +z2 2

Fig. 1.19 (a)

= ��

��

�

�

�

��

��

�

�

�

1

h z

1

h z

1

h z

1

h z222 2

12 2

22 2

12

–

Using I1 and I2in equation (1),

� E =�

��

�

�

�

�

�� &

��&��

L

0

2

222

1

212

L

0 24

z

h h z

z

h h z4

1

ha y

��

�

�

�

�

z

1

h z22 2

12

a V / mz

Example 1.6.14

Solution : Q = 1 �C and placed between A(0, 0, 1) and B(0, 0, 2) m.

� L = 2 – 1 = 1 m

� &L

=QL

=11

= 1 �C/m

Consider an elementary charge dQ at a distance

z as shown in the Fig. 1.20.

� dQ = &L

dz

i) For point P1(0, 0, 0),

R = �za z , a R = �a z

| |R = z

� dE =dQ

4 R02��

a R =&

��L

02

dz

4 z( )�a z

� E =

z 1

z 2L

02

dz

4 z( )

�

�

' �&

��a z

=�

�'

&

��L

0 z 1

2

24dz

za z

= ��

� ��

��

�

�

1 10

4 10

6

12� 8.854

1z 1

2

a z = 8987.742412��

��

1 a z

= – 4493.8712 a z V/m

ii) For point P2(0, 1, 1)

R = 0 (1 0) (1 z)a a ax y z� � � � , | |R = 1 (1 z) 2� �

� dE =dQ

4 R02��

a R =dQ

4 R02��

R|R |

=&

��L

02

dz

4

[ + (1 z) ]

1+ (1 z)[ ( ) ]1 1 2� �

�

�z

a ay z



(0, 0, z)

(0, 0, 2)

(0, 0, 1)

P (0, 1, 1)2

P (0, 0, 0)1

y

z

x

z

B

A

dz

Fig. 1.20

� dE =1 10

4

6�

� � � ��

��

�� 8.854 10

dz

[1 (1 z) ]

(1 z) dz

[112 2 3/ 2

a ay z

� �

��/

�/

��/

�/(1 z) ]2 3/ 2

� E = dE' = 8987.7424

z 1

2

2 3/ 2 2 3/

dz

[1 (1 z) ]

(1 z) dz

[1 (1 z) ]�'

� ��

�

� �

a ay z2

��/

�/

��/

�/

I 1 =

z 1

2

2 3/ 2

dz

[1 (1 z) ]�'

� �0 put 1 – z = tan ), – dz = sec2 ) d)

For z = 1, )1 = 0" and z = 2, )2 = – 45"

� I1 =

)

)) )

)1

2 2

3/ 2

sec d

[1 ]'

�

� tan 2= � '

)

)

))

1

2 1d

sec= � '

)

)

) )

1

2

dcos

� I1 = �"

� "[sin ])

0

45= � � "[sin ( )]45 = + 0.7071

I2 =

z 1

2

�'

(1 z) dz

[1 (1 z) ]2 3/ 2

�

� �0 put [1 + (1 – z)2] = u2

�2(1 – z) (– dz)=2u du i.e. (1 – z)dz = –u du

for z = 1, u1 = 1 and z = 2, u2 = 2

� I2 =

u 1

u 2

3

u du

u'

�= � ��

��

1u 1

2

=1

21��

�

� = – 0.2928

� E = 8987.7424 [0.7071 a y – 0.2928 a z ] = 6355.2326 a y – 2631.6109 a z V/m

Example 1.7.3

Solution : Consider the ring in x-y plane with uniform charge density &L . Let point

charge is at P, which is at a distance z on the axis of the ring, as shown in the Fig. 1.21.

Consider cylindrical system.



z

y

x

R

P(0, 0, z)

z

r

dQ

Q1

PL

Rz za

P

– rar

0

dQ

Fig. 1.21

R = � �r za ar z

a R =� �

�

r z

r z2 2

a ar z

dQ = &L dl = & (L ( )r d

Hence the force on Q 1 due to dQ is,

FP =Q dQ1

04�� ( )r z

r z

r z2 2 2 2�

� �

�

�

�

�

'a ar z

Due to symmetry, radial components cancel.

� FP =

(

� & (

��

&

� ��'

��

�0

21

03 2

1

03 24 4

Q r d z Q rL La z

( ) ( )[

r z r z2 2z

2 2( �]

02 a z

� FP =Q r z1

03 22

&

�L

2 2r z( )�a z

Now if point P is in the plane of the ring then z = 0.

� FP = 0 if P is in the plane of the ring …Proved

Example 1.8.5

Solution : The plane is shown in the Fig. 1.22

Consider the differential surface area dS carrying

charge dQ.

� dQ = & S dS where dS = dxdy

� dQ = � �2 x y 92 2� � 3/2 dx dy nC

� dE =dQ

4 Ro2� �

a R

R = � � � � � �� 0 – x 0 – y 0 – –3a a ax y z� �

R = –x – y 3a a ax y z� ,

| |R = x y 92 2� � , a R =RR| |

� dE =� �

� �� 2 x y 9 dx dy

4 x y 9

–x – y 3

x y

2 2 3/ 2

o2 2 2 2

� �

� �

�

� ��

� �

a a ax y z

� �9�10 9–

Due to symmetrical distribution, x and y components of dE will cancel each other and

only z component will exist.



z

y

x

O

P Q (2,2,–3)

S(–2, 2, –3)

R

z = –3 plane

(2,–2,–3)

dS

Fig. 1.22

� dE =6 10

4dx dy

–9

o

a z ��

� E =6 104

dx dy6 104

[x]–9

ox – 2

2

y = – 2

2 –9

o–

��

�

�'' � � � �

a z 22

–22[y] a z = 862.82 a z V/m.

Example 1.8.6

Solution : The sheets are shown in the

Fig. 1.23.

E =&�S

2 0a N

i) PA = (2, 5, – 5)

It is below the plane z = – 4.

Hence a N for this point due to all the

sheets is –a z .

� Et = � � � � � �&�

&�

&�

S1 S2 S3

2 2 20 0 0–a –a –az z z� � = –56.47 a z V/m

ii) PB = (4, 2, –3)

It is above z = – 4 and below other two plane. Hence a aN z� � for & S1 and –a z for & S2

and &S3 .

� Et = � � � ��

� �3 102

6 102

8 10

2

9

0

9

0

9

0

��

��

�– – ––

� � �a –a –az z z = 282.358 a z V/m

iii) PC = (–1, –5, 2)

It is above z = 1 and below z = 4. Hence a N = �a z for & S1 and & S2 while –a z for & S3 .

� Et =� �

� �3 102

6 102

8 10

2

9

0

9

0

9

0

��

��

�– – ––

� � �a a –az z z = 960.018 a z V/m

iv) PD = (–2, 4, 5)

It is above all the planes hence a N = � a z for all.

� Et =� �

� �3 102

6 102

8 10

2

9

0

9

0

9

0

��

��

�– – ––

� � �a a az z z = + 56.47 a z V/m



x

z

yz = –4&S1 = 3 nC/m

2

&S2 = 6 nC/m2

&S3 = –8 nC/m2 z = 4

z = 1

Fig. 1.23

Example 1.8.7

Solution : The plane is shown in the Fig. 1.24. The plane can be defined uniquely from

three points which can be obtained from the equation of plane x – 2y + 3z = 4.

For x = 0, y = 0, z =43

� P 0 043

, ,��

��

For x = 0, z = 0, y = – 2 � Q (0, – 2, 0)

For y = 0, z = 0, x = 4 � (4, 0, 0)

The three poionts P, Q and R define a plane.

The plane is infinite sheet of charge.

� E =&�S

2 0a n =

12

20

0

��

a n = 6 a n V/m

a n = Unit vector normal to the plane.

Note If plane is defined as Ax + By + Cz = D then the unit vector normal to the

plane is,

a n = %� �

� �

�

�

�

A B C

A B C2 2 2

a a ax y z

Positive sign for front side of the plane and negative sign for back side of the plane.

In this case, A = 1, B = – 2, C = 3, D = 4

� a n = %� �

� �

�

�

�

a a ax y z2 3

2 312 2 2= % [0.2672 a x – 0.5345 a y + 0.8017 a z ]

The origin is on the back side of the plane so use negative sign.

� E = 6 [– 0.2672 a x – 0.5345 a y + 0.8017 a z ] = – 1.6035 a x – 3.207 a y – 4.8102 a z V/m

Example 1.8.8

Solution : i) For Q = 12 nC at P(2, 0, 6)

RPO = – 2 a x – 6 a z , |RPO| = 40

� E1 =Q

4 RPO2��0

a RPO

=12 10

4 40

2 6

40

9�

� � �

� ��

�

� �

�� 8.854 10 12

a ax z

= – 0.8526 a x – 2.558 a z V/m



P

Q

R

O

y

z

xan

& �s 0= 12

Fig. 1.24

ii) &L is parallel to z-axis.

Any point on line charge is (– 2, 3, z)

� r = [0 – (– 2) a x + [0 – 3] a y

= 2 a x – 3 a y , |r| = 13

The z co-ordinate is not considered as line charge

is parallel to z-axis.

� E2 =&��

L

r2 0a r

=3 10

2 13

9�

� � �

��

�

�

�

�� 8.854 10

2 3

1312

a ax y

= 8.2963 a x – 12.445 a y V/m

iii) &S is at x = 2 and origin is on the back side

of the sheet. Hence a n = – a x .

� E3 =&�S

2 0a n

=0 2 10

2

9.( )

�

� ��

�

�8.854 10 12a x = – 11.2943 a x V/m

� E at origin = E E E1 2 3� � = – 3.8506 a x – 12.445 a y – 2.558 a z V/m

��



y

x

z

12 nC(2,0,6)

O

x = –2,y = 3

= 3 nC/m�L

–ax

Origin onback side

x = 2

= 0.2 nC/m�s

2

P

Fig. 1.25

Solutions of Examples for Practice

Example 2.5.4

Solution : a) A point charge of 40 mC at the origin.

P (6, 8, – 10) and O (0, 0, 0)

� r = ( ) ( ) ( )6 0 8 0 10 0� � � � � �a a ax y z

= 6 8 10a a ax y z� �

� r = ( ) ( ) ( )6 8 10 2002 2 2� � � �

� a r =6 8 10

200

a a ax y z� �

� D =

Q

4 r 2� �a

a a ar

x y z�

�

�

� ��

��

�40 10

4 200

6 8 10

200

3

2( )

= 6.752 10 a 9.003 10 a 11.254 10 a C m6x

6y

6z

2� � � � ��

b) � �L 40 C m� along z-axis

The charge is infinite hence,

E =��

L

02 ra r

As the charge is along z-axis there can not be any component of E along z-direction.

Consider a point on the line charge (0, 0, z) and P (6, 8, – 10). But while obtaining r do

not consider z co-ordinate, as E and D have no a z component.� r = ( ) ( )6 0 8 0 6 8� � � � �a a a ax y x y

� r = ( ) ( )6 8 102 2� �

� a r =6 8

10

a ax y�


(2 - 1)

2 Electric Flux Density and Gauss’s Law

yar

r

x

z

QO(0,0,0)

P(6,8,–10)

Fig. 2.1

� E =�

��L

02 ( )10

6 8

10

a ax y��

��

�

��

� D = ��0

L

2E

a ax y�

�

��

��

�

��10

6 8

10= 3.819 10 a 5.092 10 a C m7

x7

y2� � ��

c) � �S2C m� 57 2. on the plane x = 12.

The sheet of charge is infinite over the plane x = 12 which is parallel to yz plane. The unit

vector normal to this plane is a an x� .

� E =��S

02a n

The point P is on the backside of the plane hence a an x� � , as shown in the Fig. 2.2.

� E = � ��S

02�a x

But D = �0 E

� D = � ��Sxa

2� = � � �28.6 10 a C m6

x2

Example 2.5.5

Solution : �S = 2 �C/m, z = 2 m plane, P (1, 1, 1).

D = � ��

�0 0

S

0

S

2 2E a an n� � �

Now at P (1, 1, 1), a an z� � as the plane is at z = 2 which is above point P.

� D at P =22

C� /m ( )2 � � �a a C/ mz z2�

Electromagnetic Theory and Transmission Lines 2 - 2 Electric Flux Density and Gauss's Law


y

x

z

6

8

Planex=12

Back sideof plane

–10–ax

P

Fig. 2.2

Example 2.5.6

Solution : i) Case 1 : Point charge Q = 6 �C at P (0, 0, 0).

While D to be obtained at A (0, 0, 4).

� r = ( ) , ( ) ,4 0 4 4 42� � � � � �a a r arr

az z r z

� D1 =Q

4 r 2� �a a 2.984 10 a C mr z

8z

2��

��

��6 10

4 4

6

2( )

Case 2 : Line Charge �L = 180 nC/m along x-axis. So any point P on the charge is

(x, 0, 0), while A (0, 0, 4). As charge is along x-axis, no component of D is along x-axis.

So do not consider x co-ordinate while obtaining r.

� r = ( ) , ,4 0 4 4� � � � �a a r arr

az z r z

As charge is infinite,

� D2 =�� L

2 ra a 7.161 10 a C mr z

9z

2��

� ��

�180 10

2 4

9

Case 3 : Uniform sheet of charge lies in z = 0 plane. So the direction normal to it is z

direction as plane is xy plane. Hence a an z� and � S = 25 nC/m2 .

As sheet is infinite,

D3 =�

S

212.5 10 C m9 2a a an z z�

��

��25 10

2

9

� D = D D D 49.501 10 a C m1 2 39

z2� � � � �

ii) The point at which D is to be obtained is now B (1, 2, 4).

Case 1 : Point charge Q = 6 �C at P (0, 0, 0).

� r = ( ) ( ) ( )1 0 2 0 4 0 2 4� � � � � � � �a a a a a ax y z x y z

� r = ( ) ( ) ( )1 2 4 212 2 2� � �

� a r =rr

a a ax y z�

� �2 4

21

� D1 =Q

4 r 2� �a

a a ar

x y z�

�

�

� ��

��

�

��

�6 10

4 21

2 4

21

6

2( )

= 4.961 10 a 9.923 10 a 1.9845 10 a9x

9y

8z� � � � �� C m 2

Case 2 : Line charge : The point on the charge is (x, 0, 0).

As charge is along x-axis, do not consider x co-ordinate.

� r = ( ) ( )2 0 4 0 2 4� � � � �a a a ay z y z ... as B ( 1, 2, 4)



� r = ( ) ( )2 4 202 2� � hence a r =rr

a ay z�

�2 4

20

� D2 =��L

2 ra r =

180 10

2 20

2 4

20

9�

�

��

��

�

��

�

�

a ay z

= 2.8647 10 a 5.7295 10 a C m9y

9z

2� � ��

Case 3 : Infinite sheet of charge in z = 0 plane.

The point B ( 1, 2, 4) is above z = 0 plane hence a an z� and D3 remains same as before.

D3 =�

S 9 22

12.5 10 C ma a an z z��

� ��

�25 10

2

9

� D = D D D1 2 3� �

= 4.961 10 a 1.2786 10 a 3.807 10 a9x

8y

8� � � � �� z

2C m

iii) Let us find the total charge enclosed by a sphere of radius 4 m.

Charge 1 : Q C1 � 6 � at the origin.

Charge 2 : The charge on that part of the line which is enclosed by the sphere. The line

charge intersects sphere at x = � 4. Hence charge on the length of 8 m is enclosed by the

sphere. This is shown in the Fig. 2.3.

� Q 2 = �L � length enclosed = 180 10 8 = 1.44 C9� ��

Charge 3 : The intersection of z = 0 plane with a sphere is a circle with radius 4 m,in xy plane.

The surface area of this circle is � r 2 .

� S = �� ( )4 2 50.2654 m2



y

z

x

�L+4

4 m

–4Intersection ofz = 0 plane withsphere

�S

z = 0 plane

Fig. 2.3

Hence the total charge enclosed is,

� Q 3 = � �S9S = 25 10 50.2654 = 1.2566 C� � ��

Hence the total charge enclosed by the sphere is,

Q total = Q Q Q 8.6966 C1 2 3� � � �

But � = Q total = Total electric flux leaving the surface of sphere

= 8.6966 �C

Example 2.5.7

Solution : �v = 10 e 20 r�C/m3, sphere radius = a

Q = �v

v

dv� =

�

�

�

�

� � ��

�

��

0

2

0

10e r dr d d20r

r 0

a2 sin

r e dr2 20 r�� = r e dr 2r e dr2 20 r 20 r� �� dr ... By parts

= re

202r

e20

dr220 r 20 r� �

�

�

��

�

��

�

�

�

!

"##� =

��

��

r e

20110

r e dr2 20 r

20 r

= � � ��

��

� � �� 120

r e110

r e dr 1 e dr dr2 20 r 20 r 20 r

= � ��

�

��

�

��

�$�

� �

�120

r e110

re

–e– 20

dr2 20 r20 r 20 r

20$

��$

$

= � � � ��

�

��

�

��

�

� ��1

20r e

110

120

re120

e20

2 20 r 20 r20 r$

$

��$

$

= � � ��

��

� � �120

r e1

200r e

14000

e2 20 r 20 r 20 r

0

a

d��

0

2

� = 2 and sin0

� � �� d = 2

� Q = 10 2 2120

a e1

200a e

14000

e1

42 20 a 20 a 20 a� � � � � ��

000C�

��

On the surface of the sphere, D =Q

4 a 2�a r

Using the expresssion of Q obtained above,

D = � � � ��

��

�

��

�� 1

2e

120

ea

1400

e

a

1

400 a20 a

20 a 20 a

2 2a r C/m2



Example 2.6.4

Solution : a) Let us obtain the surface area of the sphere bounded by the given

coordinates. The differential surface area normal to a r direction is r d d2 sin � � � . The flux

is directed radially outward along a r direction.

� S =

�� %

� & '

�

�

� � �� 0

r d d2 sin and r = 10 m

= � � ( ) ( ) � �10 100 220 0� � � �cos � �

�'

� � & ' = 314.1592 m2

Total flux passing through the total surface area is,

� = Q = 25 �C through S r mtotal2 2�4 � which is surface area of total sphere.

Hence flux passing through given portion of the sphere i.e. S is,

� =� �

314.1592 25 10

4 10

6

2

� �

�

�

�= 6.25 �C

b) The closed surface is shown in the Fig. 2.4.

The point charge is completely enclosed by the closed surface. According to Gauss's law,

the total flux passing through any closed surface is charge enclosed hence,

� = Q = 25 �Cc) The plane is shown in the Fig. 2.5.



Q

z = –0.5

y

x

r = 0.8 mz = +0.5

z

Fig. 2.4

As the plane is infinite, half the total flux originating from charge will pass through the

plane.

� � =Q2

252

� = 12.5 �C ... As � total = Qtotal

Example 2.6.5

Solution : The charges are

shown in the Fig. 2.6.

Consider line charge along

x-axis. Any point Q on this

charge is (x, 0, 0). As the

charge is infinite along x

axis, E and hence D has no

component in a x direction.

� Q (x, 0, 0) and P (3, 3, 3)

� r = � � � �3 0 3 0� � �a ay z... x co-ordinate need not beconsidered.

� r = 3 3a ay z� and |r|� � �9 9 18

� D1 =�� L

2 ra

a ar

y z�

�

�

��

��

�

��

�25 10

2 18

3 3

18

6

= 6.6314 � �10 7 a y + 6.6314 � �10 7 a z C/ m2

Consider any point Q on charge along y axis.



Qy

x

z

Plane z=6

Flux

Fig. 2.5

P(3,3,3)

�L�L

x

y

z

Fig. 2.6

Hence Q (0, y, 0) and P (3, 3, 3). There is no component of E hence D along a y direction

as charge is along y axis. So do not consider y co-ordinate.

� r = � � � �3 3a ax z� and r � � �9 9 18

� D2 =�� L

2 ra

a ar

x z��

�

��

��

��25 10

2 18

3 3

18

6

= 6.6314 � �10 7 a x + 6.6314 � �10 7 a z C/ m2

Hence total D at point P due to both the charges is,

D = D D1 2�

= 0.6631 a x + 0.6631 a y + 1.3262 a z �C/ m 2

Example 2.6.6

Solution : a) The flux leaving is charge enclosed.

� = Q =

SS

r 0

5

2dS

5r

r 1r dr d� � ��

��

� �� %

'�

The dS = r dr d� as the �S is in plane z = 2, to which the normal direction is a z , as

shown in the Fig. 2.7.

� � =

�� %

'�

�� r 0

5 2

2

5 r

r 1dr d

Now ��

x dx

a x c

2

2=

xa

ca

1

actan x

ac

1��

�

!

"##

�

��

�

��

�

� � = ( ) ( )� �502� � r

1tan r1

0

5

��

��

� ... a = c = 1



dS

az

r<5

z=2plane

�S

x

y

z

Fig. 2.7

= ( )5 2 5 51� � � �� tan = 113.932 nC ... use radian mode

b) Half of the flux leaves in a z direction while other half leaves in �a z direction.

� � leaving in �a z direction =113.932

2= 56.966 nC

Example 2.6.7

Solution : D a ar2 r z cos r z sin cos r cos2 2 2� � �� a z

i) r = 3, 0 z 5* *

The surface is cylindrical as shown in the Fig. 2.8.

� Q =

S top bottom side

d d� � � �+ +� � �D S D S

For top, z = 5, d r dr dS a z� �

� D Sd+ = r cos dr d3 2 � � ... a az z+ = 1

�top

d� +D S =

�

'�

� ��

0 r 0

33 2r cos dr d

=( )

�

� ��

0

2

r 0

33r

1 cos 2

2dr d

� ��

�

=r4

12

4

0

3�

��

�

��

� � ��

��

��

�

��

�

� �sin 2

20

2

= 63.617 C

For bottom, z = 0, d r dr dS a z� � �

� D Sd+ = � r cos dr d3 2� � i.e.

bottom

d� +D S = – 63.617 C

For lateral surface, r = 3, d r d dzS a r� �

� D Sd+ = 2 r z cos d dz2 2 � � ... a ar r 1+ �

�side

d� +D S =

�

�

�

�

� ��

0

2

0

2

z 0

52 2 2

z 0

5

2 r z cos d dz 2 r( )

z1 cos 2

2d dz

� ��

=2 r

2z2

2 2

0

5�

��

�

��

��

��

��

�

��

�

� �sin 2

20

2

... r = 3 constant

=2 9

2252

2�

� � � = 706.858 C

� Total flux = 63.617 63.617 706.858� � �706.858 C



z = 5

z

r=3

z = 0

Fig. 2.8

ii) z = 0, 0 r 3* *

This surface is circle in xy plane as shown in the Fig. 2.9.

This is nothing but bottom surface of the cylinder considered in

part (i) above.

� Electric flux = – 63.617 C

Example 2.7.6

Solution : a) At r = 2 cm, it is inner side of inner sphere. It is seen that inside a

spherical shell with surface charge E and D = 0. Now r = 2 cm is inside of all three

spheres hence E D= = 0.

At r = 4 cm which is exterior to innermost sphere but inside of spheres having radii 5and 7 cm. Hence at r = 4 cm, D and E exist due to sphere of r = 3 cm with

� �S � 200 C/m2 .

E =�

�S

2

02

a

ra r

=� ��

200 10 3 10

8.853 10 4 10

6 2 2

12 2 2

� � �

� � ��

� �

� �a 12.706 10r

6 a r V/m

Here a = Radius of sphere = 3 cm and r = 4 cm is distance.

and D = � �0 E 112.5 a C mr2�

At r = 6 cm, the E and D will be due to the two spherical shells having radii 3 and 5

cm. While due to sphere of r = 7 cm, D and E are zero at r = 6 cm.

� a 1 = 3 cm , � �S12C m� 200

� E1 =� ��

�S1 1

2

02

6 2 2

12

(a )

(r)

200 10 3 10

8.854 10 6 1

a r ��

� � �

� �

� � �0 2 2�a r = 5.6471 10 a6

r� V/m

� D1 = � �0250 C mE a r�

And a 2 = 5 cm, � �S2250 C m� �

� E2 =� ��

�S2 2

2

02

6 2 2

12

(a )

(r)

10 5 10

8.854 10 6 1

a r ��

� � �

� �

�

50

� �0 2 2�a r = � �3 106.9216 a r V/m

� D2 = � �0234.722 C mE a r� �

� E = E E 1.7255 10 a1 26

r� � � V/m

and D = D D 15.278 a C m1 2 r2� � �



z

z = 0

r = 3

0

x

y

Fig. 2.9

Note that radial distance r is measured from the centre i.e. origin of the spheres.

b) The spheres are shown in the Fig. 2.10.

At r = 7.32 cm, all three shells produce D.

� D1 =� S1 1

2

2

(a )

(r)a r , D2 =

� S2 22

2

(a )

(r)a r , D3 =

� x 32

2

(a )

(r)a r

But D = 0 at r = 7.32 cm as given.

� D = D D D1 2 3� � � 0 i.e.� � �S1 1

2S2 2

2x 3

2

2

(a ) (a ) (a )

(r)

� ��a r 0

But r , 0 and a r , 0

� � � �S1 12

S2 22

x 32(a ) (a ) (a )� � = 0

� � x =� � � � � �

� ��

� � � � � � � �

�

�

�

��

�

�

��

� � � �

�

200 10 3 10 50 10 5 10

7 10

6 2 2 6 2 2

2 2�

= – 11.2244 �C m 2

Example 2.7.7


The spherical surface A shown in the Fig. 2.11 (b) is the Gaussian surface for the line

charge. Let the differential surface area is dS = r d dz� to which a r is normal. The D is

directed radially outwards. The length of the Gaussian surface is L.

� D = Dr a r and d r d dzS a r� �The radius r of Gaussian surface A is 0 < r < 3.



r = 7.32 cm

a = 7 cm3

a = 5 cm2

a = 3 cm1

Fig. 2.10

� Q =

S S

D S� �+d = D r d dzr � ... � �a ar r+ �1

=

z 0

L

0

2

r r 0LD r d dz = D r z]

� ��

�

�� )[ [

02 = D r 2 Lr �

But charge on the line of length L is Q = �L1 L�� L1 L = D r 2 Lr �

� Dr =��

��

L1 L1

2 r=

2 rand D a r

� D =2.5 10

2 r

6��

�

�a r

0.3978r

a C mr2� for 0 < r < 3 m

The spherical surface B is the Gaussian surface enclosing both the charge distributions.

Due to the line charge, D1 =��L1

2 ra r remains same.

And due to cylinder of radius 3 m, let it be D2 . The direction of D2 is radially outwards.

Consider differential surface area normal to a r which is r d dz.� The length of Gaussian

surface is L.

� D2 = D d r d dz2r a S ar rand � �

� Q =

S2D S� � �+ �

� �

d D r d dz

z 0

L

0

2

2r�

�

� = D r 2 L2r � ... � �a ar r+ �1

Now charge on the surface of length L and radius r is,

Q = � S � Surface area = � �S 2 rL�where r = 3 m = Radius of charge distribution



�S

r = 3m

�L1

z

z�L1

A

B

r = 3m

�S

(a) Charge distribution (b) Gaussian surfaces

Fig. 2.11

= � �2� � � � � � � � �� 0.12 10 (3) L 2.2619 10 L C6 6

� � � �2.2619 10 L6 = D r 2 L2r � � � i.e. D2r =� �

��

��

�2.2619 102 r

0.36r

6

�10 6

� D2 =� 0.36

rC m2a r � for r > 3

� D = D D1 2� = 0.0378r

a C mr2� for r > 3

Example 2.7.8

Solution : The charge distribution is shown in the

Fig. 2.12.

a) For r * R

Gaussian surface is for r < R.

Q =

�

�

�

�

�v

r

r

dv

� � ��

0

2

0 0

=

�

�

�

� �� 2

r

r r

Rr dr d d

� � ��

0

2

0

0

0

sin = ( ) ( )�

� �� 0

00 0

2

Rr4

4 r�

��

�

��

– cos =�� 0 r

R

4

D =Q

r 24�a r =

��

�0

4

r

R r

4

2�a r =

� 0

4

r

R

2

a r

� E =D�0

=��0

2

0r

r4 R

a … for 0 r R* *

b) For r > R

Gaussian surface is for r > R.

� Q =

�

�

�

�

�v

r

r

dv

� � ��

0

2

0 0

=

�

�

�

�

�

�

�

�v

r

R

dv

� � � � ��

0

2

0 0 0

2

0

�

��

-

v

r R

dv

=

�

�

�

� �� 2

r

R r

Rr dr d d

� � ��

0

2

0

0

0

0sin

= ( ) ( )�

� �� 0

00 0

2

Rr4

4 R�

��

�

��

– cos = � �0 R 3

� D =Q

r 24�a r =

� �

�0

4

R

r

3

2a r =

� 0

4

R

r

3

2a r

� E =D�0

=�

�0

3

02 r

R

4 ra … for r > R



r < Rr > R

R

0

+

++

+

Fig. 2.12

Example 2.7.9

Solution : a) E =D

a r� �0 0

r3

� and r = 0.2 m

� E =0.2 10

3 8.854 10

9

12

�

� ��

�

�7.5295 a r V/m

b) Q = � �S

d� +D S

Consider a differential area dS normal to a r which is r sin d d2 � � �.

�dS = r sin d d2 � � � a r and D =r3

a r

�D S+d = � �r3

sin d d3

� � � ... a ar r+ �1

� Q = ( ) ( )�

�

�

��

= 0

2

= 0

3 3r3

sin d dr3� � �cos

0 02

=43� r nC3 ... Note that D is in nC/m 3

�At r = 0.2 m,Q =43�� (0.2) 3 = 0.0335 nC = 33.51 pC

c) At r = 0.3 m,Q =43�� (0.3) 3 = 0.113 nC = 113.097 pC

Example 2.7.10

Solution : i) The spherical surface at r = 5 encloses all the shells with r1 = 1, r2 = 2

and r3 = 3.

� Q = Charge enclosed by surface at r = 5

= Charge enclosed by surfaces (r1 + r2 + r3)

= Q1 + Q2 + Q3 = � � � � � �S S Sr r r1 12

2 22

3 324 4 4� � � � �

= 4� [20 � 10 – 9 � 12 – 9 � 10– 9 � 22 + 2 � 10– 9 � 32] = 25.1327 nC

� Flux leaving the surface at (r = 5) = Q = 25.1327 nC

ii) P(1, – 1, 2) is in cartesian form

� r = ( ) ( ) ( )1 1 22 2 2� � � = 2.4494 m

Hence shells r1 = 1 and r2 = 2 are enclosed. Where D =� S

r

a 2

2a r for a spherical shell with

radius a at r > a.

� D at P = D1 due to (r1 = 1) + D2 due to (r2 = 2) while D at P = 0 due to r3 = 3 as

r = 2.4494 is inside the shell.



� D at P =� �S S

r r

12

22

(a (a12

22)

( )

)

( )a ar r�

=20 10 (1)

(2.4494)

( 9 10 ) (2)

(2.4494)

9 2

2

9 2

2

� ��

� � ��

��

��

��

a r = – 2.667 a r nC/m2

Example 2.7.11

Solution : a) To find Q tot use standard result as �v is constant.

Q tot =

v

v� � dv = � �43

� �r 3v

r 10 cm�... � �

v

3dv r� �43�

= � �43

� 0.1 43 � = 0.016755 �C

Alternatively, Q tot =

�

�

�

'� .

� � � ��

0 0

v

r 0

0 1 m2r dr d dsin = 0.016755 �C

b) To find Dr , consider a Gaussian

surface as a sphere of radius r as

shown in the Fig. 2.13. Consider dS at

point P. The D is in a r direction hence

D a r� Dr and dS normal to a r is

r d d2 sin � � �.

� dS = r sin d d2 � � � a r

� Q =

S

r2d D r d d� � �+ �

�

D S�� %

�

�

�

� � �2

0

sin ... � �a ar r+ �1

� Q = ( ) ( )D rr2 �cos � ��

0 02

� Dr =Q

r 24 �and D a r�

Q

r 24 �

But Q =43

� �r 3v for a sphere of r

� D =

43

4 10

4

6�

�

r

r

3

2

� �

= 1.333 r �C/ m 2

c) Let charge between 10 cm < r < / is Q 1.

� Q 1 =

v

v� � dv =

�� %

'�

�

� /

� � � �� 0 r 0.1

v2r dr d dsin = ( ) ( )�

�

��cos � �� /

0 02

r 0.1

2

3

3 r

r 0.001dr



D

ar

dS

rr = 10 cm

Gaussiansurface

P

Fig. 2.13

Put r 3 + 0.001 = u i.e. 3 r dr2 = du

� Q 1 = ( )2 2 4� � �/

/� � � ��

��r 0.1

r 0.1

duu

uln

Resubstitute u = r 3 + 0.001,

� Q 1 = ( )� � ��

�

�

��

�

��

4 42 10 3

� �// 0

ln lnr 0.0010.0013

0.1nC

Hence the total charge for 0 < r < / is, Q Qtot 1� i.e. resultant charge Q R is

Q R = 0.016755 � ��

�

�

��

�

��

��10 4

2 10106

39�

/0ln

0.001C

But required Q R = 0

� 410

103

9�/0

ln�

�

�

��

�

��

��0.001

2= 0.016755 � �10 6 i.e. ln

/ 3

3

0.001

2 10

�

�

�

��

�

��

= 1.3333

�/ 3

3

0.001

2 10

�

� �= e1.3333 = 3.7936 i.e. /0 = 6.5872 � �10 3

� / = 0.1874 m = 18.74 cm

Example 2.7.12

Solution : The region is shown in the Fig. 2.14.

a) Region r < 2 m

For this region, there is no charge enclosed hence D 0� .



-

– -

2 m

� C/m3

4 m

z

Dotted shown isGaussian surface

(a) (b)

L

Fig. 2.14

b) Region 2 < r < 4 m

Consider Gaussian surface in cylindrical form of height L and radius r such that 2 < r < 4

as shown in the Fig. 2.14 (b)

D = Dr a r ... D is in radial direction.

d S = r d dz� a r

Q =

S z 0

L

0

2

rd r dr d D� � �+ ��

D S�

�

� = r D 2 Lr � ... a ar r+ � 1

But Q enclosed by Gaussian surface at 2 < r < 4 is,

Q = �� [Volume of Gaussian surface at 2 < r < 4] = � ��

�

z 0

L

0

2

r 2

r

r dr d dz

� � ��

= � �( )� � �� r 2 L2 2 = � �� r 4 L2 � �

� � �� r 4 L2 � � = r D 2 Lr � i.e. Dr =� �� r 4

2 r

2 �

� D =� �� r 4

2 ra C/ m

2

r2

�... 2 < r < 4

c) Region r > 4 m

Again Q = r D 2 Lr �But Q enclosed by Gaussian surface is Q enclosed by the entire cylindrical region of length

L as r > 4 m.

� Q = �� Volume enclosed = � ��

�

z 0

L

0

2

r 2

4

r dr d dz

� � ��

= ( ) ( )1 2� � �� 4 2 L2 2 = � �� 16 4 L 12 L� �

� 12� � L = r D 2 Lr �

� Dr =6r�

i.e. D =6r

a C/ mr2�

... r > 4

Example 2.9.6

Solution : Use Gauss's law in point form, 3+D = �v

Given D in spherical coordinates hence,

3+D = � � � �1

r rr D

rD

r

D

22

r44

�44

�4

41 1

sinsin

sin� ��

� ��

…Spherical

Now Dr = 10 sin �, D� = 2 cos �, D� = 0



� 3+D = � � � �1

r rr

r224

4�

44

�101

2 0sinsin

sin cos��

� �

= � � � �1

r rr

r2210

12sin

sinsin�

� ��

44

�44

= � � ( )10 1

2 2sin

sincos

��

�r

2rr2

� =20 2sin cos

sin

� ��r

2r

�

� �v =20 2sin cos

sin

� ��r

2r

C m 3�

Now cos 2 � = cos sin2 2��

� �v =( )20 2 2 2

sin cos sin

sin

� ��

�r r�

=20 2 182 2 2 2 2sin cos sin

sin

sin cos

sin

� ��' ��

�� ' ��

��

r r

= ( )sin cos

sin

� �

�

��

r18

2 2

2�

�

��

�

��

� �sin

r18 2 cot C/ m2 3

Example 2.9.7

Solution : i) The surface x = 2, 0 y 2* * and

0 z 2* * is shown in the Fig. 2.15.

Key Point The plane is parallel to yz plane and

unit vector normal to the plane is a x .

The direction away from the origin is a x and

dS = dy dz.

� dS = dy dz a x

� D S+d = 2y z dy dz2 2

as a ax x+ = 1 and a a a ay x z x+ +� � 0

� Q =

S� +D Sd

= 2y z dy dz

z= 0

2

y= 0

22 2� �

= 2y

3z3

2 83

83

3

0

23

0

2�

��

�

��

�

��

�

��

��

� � 14.22 pC

ii) r = 1 �m for the sphere. So incremental volume of the sphere is,



x=y=z=0

x

x=2 x=2

ax

y=2

z=2

z

y

Planebounded by

0 y 2* ** *0 z 2x = 2

Fig. 2.15

dv = � �43

r43

1 10 4.1887 10 m3 3 18 3� �� 6

� dQ = charge contained in volume dv = �v dv

Now �v = ( )3 �44

�4

4�

44

�

��

�

��+D

at P (2, 2, 2)x y zD

x

D

yDz

( , , )2 2 2

= ( )0 6xyz 2xy2

(2, 2, 2)� � = 96 + 8 = 104 pC/m 3

� dQ = 104 10 4.1887 1012 18� � � � �� 4.3562 10 16 pC

Example 2.9.8

Solution : i) �v = 3 �44

�4

4�

44+D

Dx

D

yDz

x y z = y x 1 C/ m2 2 3� �

ii) Assume given D in spherical co-ordinate system.

�v = � � � �3 �44

�44

�+ 1

r rr D

1r sin

sin D1

r22

rD� �

� � sin

D

� ��4

4

= ( )1

r rr 0 0

1

r2r

22

2

44

� � � � �2r

C/ m3

Example 2.9.9

Solution : D = � � � �2� � � � � � �� z 1 cos z 1 sin cos C m2� � �a a a z–

i) �v = 3 5 D = ( )1 1�

44�

��

4

4 �44�

�D

D Dzz� �

= ( ) ( ) ( )12

1�

44�

� � �6� ��

44�

� �6� �44

� �� ( cos – (z z sinz

cos

= ( ) ( )41

0z z cos�6 ��

� �6 �cos –� � � = 3[z 1] cos C m3� �

ii) Q =

v

v dv� � � �z� � ��

0

4

0

2

0

2

�

�

�

� � � �3 z 1 cos d d dz

= ( )32 2

2

0

2

02

2

0

4�

� ��

��

�

��

��

��

�

��

sinz

z = 3 2 1 12� � � = 72 C�

iii) Q = D S5� d

S

= D S D S D S D S D S D S5 � 5 � 5 � 5 � 5 � 5� � � � �� d d d d d d

z� � � � �0 2 0 2 0 z��

4

D S5�� d

� 0

= � �( )( )2 z 1 cos d dz� � � �� = 0



D S5�� d

� 2

= � �( ) � �( )z

z cos d dz

� ��

0

4

0

2

2 1

�

�

� � � � ...� � 2

= � � ( )2 22

20

22

0

4

� ��

��

�

��

sinz

z� �= 96 C.�

D S5�� d

� 0

= � �( )( )– –� � �z sin d dz�� 1 = � �z� ��

0

4

0

2

1

�

� � �z sin d dz = 0

D S5�� d

� � 2

= � �z� ��

0

4

0

2

1

�

� � �– z sin d dz = – sin2

zz

� ��

!"#�

��

�

��

��

��

�

��

2

0

22

0

4

2 2= –24 C� .

D S5�� d

z 0

= � �� cos d d

z

–

��

0

=

�

�

�

� � � ��

0

2

0

22– cos d d

= ( )–�

� �3

0

2

02

3

�

��

�

��

sin = –83

C�

D S5�� d

z 4

=

�

�

�

� � � ��

0

2

0

22 cos d d = ( )�

� �3

0

2

02

3

�

��

�

��

sin = +83

C�

� Q = 0 96 0 2483

83

� � �– – = 72 C.� ...Gauss's law is verified.

Example 2.9.10

Solution : D = 4 x3 a x – 2z a y – 2y a z

i) 3 5 D =44

4

444

Dx

D

yDz

x y z� � = 12 x2 C/m3

ii) �v = 3 5 D = 12 x2 C/m3

iii) Q = �v

vol� dv =

z 1

1

y 1

1

x 1

1

� � � � � �� 12 x2 dx dy dz

= 12x

y z3

1

1

11

11

3

�

��

�

��

��

� �[ ] [ ] = 12 �23

� 2 � 2 = 32 C

iv) To find Q without finding �v .

� Q = D Sd5�S

Consider all faces of the region – 1 < x, y, z < 1.



1) For front surface, x = + 1, dS = dy dz a x

2) For back surface, x = – 1, dS = – dy dz a x

3) For right surface, y = + 1, dS = dx dz a y

4) For left surface, y = – 1, dS = – dx dz a y

5) For top surface, z = + 1, dS = dx dy a z

6) For bottom surface, z = – 1, dS = – dx dy a zFor front, D Sd5 = 4x3 dy dz = 4 dy dz … x = 1

For back, D Sd5 = – 4x3 dy dz = 4 dy dz … x = – 1

For right D Sd5 = – 2z dx dz

For left D Sd5 = 2z dx dz

For top D Sd5 = – 2y dx dy

For bottom D Sd5 = 2y dx dy

D Sd5�S

=

z 1

1

y 1

1

(front)

� � � �� 4 dy dz +

z 1

1

y 1

1

(back)

� � � �� 4 dy dz +

z 1

1

x 1

1

(right)

� � � �� – 2 z dx dz

+

z 1

1

x 1

1

(left)

� � � �� 2z dx dz +

y 1

1

x 1

1

(top)

� � � �� – 2y dx dy +

y 1

1

x 1

1

(bottom)

� � � �� 2 y dx dy

� Q = 4 [y] [z] 4[y] [z] 2z2

[x] 21

11

11

11

12

1

1

11

� � � ��

��

��

�

��

�z2

[x]2

1

1

11�

��

�

��

�

– 2y

2[x] 2

y

2[x]

2

1

1

11

2

1

1

11

�

��

�

��

��

��

�

��

��

�



z

y

x

(a) –1 < x,y,z<1

–ax back

+axfront

–ay

left right

+az

top

+ay

–az

bottom

(b) Directions of dS

Fig. 2.16

= 4 � 2 � 2 + 4 � 2 � 2 + 0 + 0 + 0 + 0 = 32 C

Example 2.9.11

Solution : a) From given D

D = � � � � � �12 x 3 z 9 y z2 2 3 2 2 2� � � �

= 144 x 9 z 81 y z4 6 2 4� �

The D is maximum, when x, y and z are maximum in the given region.

� x = y = z = 2 ... maximum values

� At P(2, 2, 2), D will be maximum.

Dmax

= 144 2 9 2 81 2 24 6 2 4� � � � � � = 89.8 C/ m 2

b) According to Gauss's law in point form,

3 5 D = �v

� 3 5 D =44

�4

4�

44

D

x

D

y

D

zx y z = 24 x + 0 – 18 yz

� �v = 24 x – 18 yz

�v will be maximum when x is minimum and yz are maximum. i.e. x = + 1 and

y = z = 2.

� �v max= � �24 1 18 2 2 24 72� � � � � � � � � 48 C/ m3

c) �v is maximum when x is maximum i.e. 2 and y, z are minimum i.e. y = z = 1. Thus

�v is maximum at P(2, 1, 1).

�v max = � �24 2 18 1 1� � � � � � 30 C/ m3

Example 2.9.12

Solution : D = r r z2sin cos� � �a a ar z� �2 3 … Cylindrical

�v = 3 5D =1 1

23

r rr r

rr

z

z

244

�44�

�4

4( sin ) ( cos )

( )� � �

=1

21

2 6r

rr

r( zsin – sin )� �� = 6z C/ m3

Example 2.9.13

Solution : i) Q = � �� v

v

2r 2dv 10 e r sin dr d d

=

�

�

�

�

0

2

0 0

r

� �� 10 e r sin dr d d2r 2� � � �



0

r2r 2e r dr� � = r e dr 2r e dr dr2 2r 2r� � �� ... By parts

=r e

22r e

2dr

r e2

r e2 2r 2r 2 2r

2� � �

��

��

��

�� r 2rdr 1 e dr dr��

=r e

2r

e2

e2

dr2 2r 2r 2r� � �

��

�

�

�

!

"## � ��

= � � ��

��

�

��

� ��r e

2r e

214

e2 2r 2r

2r

0

r

= � � � ��

�r e2

r e2

14

e14

2 2r 2r2r

� Q = 100

� � � ��

��

�

��

��

�r e2

r e2

14

e14

2 2r 2r2r [ cos ] [� �� ]

02�

= 40 � � � � ��

��

�

��

� ��r e

2r e

214

e14

C2 2r 2r

2r

� D =Q

4 r 2�a 10

e2

e2 r

e

4 rr

2r 2r 2r

2� � � �

� � � 1

4 ra C/ m

2 r2�

�

��

�

��

ii) 3 5 D =1

r rr D

22

r44

... D only in r direction

= 1 21

r r5 r e 5 r e 2.5 e 2.5

22 2r 2r 2r4

4� � � ��

=1

r10 r e 10 r e 5 e 10 r e 5

22r 2 2r 2r 2r� � � � �� 1 2e 02r� �

= � � � � ��

�� 10 e

r10 e

5 e

r

10 er

5 e2r2r

2r

2

2r 2r

22r

vr

10 e� ��

Hence the result obtained is correct.

Example 2.9.14

Solution : According to Gauss’s Law,

Q =

v

( ) dv� 3+D and dQ = ( ) dv3+D

Here, dv = 10 9� m 3 at origin i.e. x = y = z = 0

3+D =44

�4

4�

44

� � � �� Dx

D

yDz

e sin y e (x y z x x sin y) 2�

� dQ = ( ) dv [0 0 2] 10at origin3 � � � � �+D � 9 = 2 10 C 2 nC9� ��



Example 2.10.4

Solution : The given D is in spherical

coordinates. The volume enclosed is shown in the

Fig. 2.17

According to divergence theorem,

S

d� +D S = � �v� 3+D dv

The given D has only radial component as given.

Hence D5 r

4r

2� while D D� �� 0.

Hence D has a value only on the surface r = 4 m.

Consider dS normal to the a r direction i.e.

r d d2 sin � � �

� dS = r d d2 sin � � � a r

�D S+d = � �r d d5r

42

2sin � � �

�

�

!

"##

=54

r d d4 sin � � � ... � �a ar r+ �1

�S

d� +D S =

�� %

'�

�

� & 7

� � �� 0

54

r d d4 sin

= ( ) ( )54

r 4 �cos � �� & 7 �0 0

2and r = 4 m

= � � � � ( )54

4 0 24 � � ��

��

cos cos�7

� = 588.896 C

To evaluate right hand side, find 3+D.

3+D = � � � �1

r rr D

rD

r

D

22

r44

�44

�4

41 1

sinsin

sin� ��

� ��

= � �1

r rr

54

rr r

r2

2 22

444

��

!"#

�

��

��

44

0 05

4= � �5

4 r4 r

23 = 5r

In spherical coordinates, dv = r dr d d2 sin � � �

� � �v� 3+D dv = � ��

�� %

'�

�

� & 7

� � �� 0r 0

425 r r dr d dsin = ( ) ( )5

0 02r

4

4

0

4�

��

�

��

�cos � �� & 7 �



z

x

y

45º

a�

D�

r = 4 m

Dr

ar

Fig. 2.17

= � �544

0 24

� � � � ��

��cos cos

�7

� = 588.896 C

Example 2.10.5

Solution : The volume defined by six planes is a cube.

i) Q =

S

d� +D S ... Surface integral

Consider all six faces of the cube as shown in the Fig. 2.18

1) Front surface (x = 3), dS = dy dz, direction = a x , d dy dzS a x�

2) Back surface (x = 1), dS = dy dz, direction = a x , d dy dzS a x� �

3) Right surface (y = 4), dS = dx dz, direction = a y , d dx dz yS a�

4) Left surface (y = 2), dS = dx dz, direction = �a y , d dx dzS a y� �

5) Top surface (z = 5), dS = dx dy, direction = a z , d dx dyS a z�

6) Bottom surface (z = 3), dS = dx dy, direction = �a z , d dx dyS a z� �

For front D S+ �d x y z dy dz3 2 x = 3 ... a ax x+ � 1

For back D S+ � �d x y z dy dz3 2 x = 1 ... a ax x+ � 1

For right D S+ �d x y z dx dz3 2 y = 4 ... a ay y+ � 1

For left D S+ � �d x y z dx dz3 2 y = 2 ... a ay y+ � 1

For top D S+ �d x y z dx dy3 2 z = 5 ... a az z 1+ �

For bottom D S+ � �d x y z dx dy3 2 z = 3 ... a az z 1+ �

�S

d� +D S =

z 3

5

y 2

43 2

z 3

5

y 2

4

3 y z dy dz 1

� � � �� 3 2y z dy dz



x = Constant planes(back and front)

y = Constant planes(sides)

z = Constant planes(top and bottom)

–ax

ax–ay ay

az

–az

Back

FrontLeft Right

Top

Bottom

(a) Cube (b) Directions of dS

Fig. 2.18

� �� (4) x z dx dz

z 3

5

x 1

32 3

z 3

5

x 1

32 3(2) x z dx dz� �

� �� (5) x y dx dy

y 2

4

x 1

33 2

y 2

4

x 1

33 2(3) x y dx dy� �

= 27y

3z2

y

3z3

2

42

3

5 3

2

4�

��

�

��

�

��

�

��

��

��

�

��

2

3

5 4

1

3 2

3

5

216

x4

z2

�

��

�

��

��

��

�

��

�

��

�

��

��

��

�

��

�

��

�

��

��

��

�

��

4x4

z2

5x4

4

1

3 2

3

5 4

1

3y

33

x4

y

3

3

2

44

1

3 3

2

4�

��

�

��

��

��

�

��

�

��

�

��

= 4032 – 149.333 + 2560 – 640 + 1866.666 – 1120 = 6549.333 �C

ii) Q =

v� 3+D

� 3+D =44

�4

4�

44

� �Dx

D

yDz

3 x y z + 2 x yz xx y z 2 2 3 3 y 2

�v� 3+D = � �

z 3

5

y 2

4

x 1

32 2 3 3 23 x y z + 2 x y z + x y dx dy dz

� � ��

=

z 3

5

y 2

4 32

4 423 x

3y z +

2 xy z +

xy dy dz

� ��

�

��

�

��4 4

... Integrating w.r.t.x

=

z 3

5

y 2

42 2[26 y z + yz + y dy dz

� �� 40 20 ]

=

z 3

5 3 2 326 y

3z +

40 y+

20 ydz

��

�

��

�

��

z

2 32

4

... Integrating w.r.t. y

=

z 3

5

[485.333 z+ 240 z+ 373.333]dz

��

=485.333 z

2+

240 z+ 373.333 z

2 2

23

5�

��

�

��

= 3882.6664 + 1920 + 746.666 = 6549.333 �C



Example 2.10.6

Solution : The given E is in spherical co-ordinates and has only radial component.

� Er =r3

E E�� , � � 0

Hence E has a value only on the surface r = a m. Consider dS normal to the a r direction

i.e. r sin d d2 � � �.

� dS = r sin d d2r� � � a

� E S+d = (r sin d d )r2 � � ��3

�

�

!

"# ... a ar r 1+ �

=

S

3d rsin d d

� � �+ ��

E S�

�

�

�

��

�0

2

03

and r = a m

=a

3[ cos ] [ ]

2 a3

23

0 0

3��

� ��

�� '�

� � � �4 a

3C

3� ��

3+E =1

r r(r E )

22

r44

... Other terms are zero as E E 0� ��

=1

r rr r

3 3 r3r

2

2

224

4��

��

�

��

� � ��

��

�

��

dv = r sin dr d d2 � � �

v

dv� 3+E =

�

�

�

� ��

� � ��

0

2

0 r 0

a2r sin dr d d

= ( )��

� ��

�� r

3a3

0

a�

��

�

��

� � � �cos [ ]0 0

23

32 2 =

4 a3

C3��

�

�S

d� +E S =

v

) dv� 3+( E

Example 2.10.7

Solution : A = x (x y ) 24(x y z )2 2 2 2 2 2a a ax y z� �

Q = A S+� d

S

The Fig. 2.19 shows unit cube centered at origin.

For A S+d , consider all six faces of the cube.

Find dS for each surface.

1) Front (x = 0.5), dS = dy dz a x

2) Back (x = –0.5), dS = dy dz (– a x )

3) Right (y = 0.5), dS = dx dz (a y )



x

y

z

1

O

1

1

Fig. 2.19

4) Left (y = – 0.5), dS = dx dz (– a y ) 5) Top (z = 0.5), dS = dx dy a z

6) Bottom (z = – 0.5), dS = dx dy (– a z )

For front, A S+d = x2 dy dz (x = 0.5)

For back, A S+d = – x2 dx dz (x = – 0.5)

For right, A S+d = x2y2 dy dz (y = 0.5)

For left, A S+d = – x2y2 dx dz (y = – 0.5)

For top, A S+d = 24 x2y2z2 dx dz (z = 0.5)

For bottom , A S+d = – 24 x2y2z2 dx dy (z = – 0.5)

� A S+� d

S

= x dy dz x dydz2

y 0.5(x = 0.5)

0.5

z 0.5

0.52

y 0.5� ��

(x = 0.5)

0.5

z 0.5

0.52 2

y 0.5(y 0.5)

0.5

x y dxdz

��

�

�� z 0.5

0.5

� ��

+ � ��

�� x y dxdz 24x y z dxdy2 2

x 0.5(y 0.5)

0.5

z 0.5

0.52 2 2

x 0.5(z 0.5)

0.5

y 0.5

0.52 2 2

x 0.5(z

24x y z dxdy

��

��

��

��

��0.5)

0.5

y 0.5

0.5

= (0.5)2 [y] [z] ( 0.5) [y] [z]0.5

0.50.5

0.5 20.5

0.50.5

0.� � � �� 5 2

3

0.5

0.5

0.50.5(0.5)

x3

[z]��

��

�

��

�

� ��

��

�

��

��

��( 0.5)

x3

[z] 24(0.5)x3

23

0.5

0.5

0.50.5 2

3

��

�

��

�

��

�

��

� ��

��

�

� �0.5

0.5 3

0.5

0.5

23y

324( 0.5)

x3 �

��

�

��

��

� �0.5

0.5

0.5

0.5y

30

Using divergence theorem,

3 5A =44

4

444

Ax

A

yA

zx y z� � = 2x + 2x2y + 48 x2y2z

( )3 5� Av

dv = [2x 2x y 48x y z]2 2 2

x 0.5

0.5

y 0.5

0.5

z 0.5

0.5

� �� dx dy dz

= x2x y

3

48x y z

3dy dz2

3 3 2

0.5

0.5

y 0.5

0.5

z

� ��

��

�

��

�

��

0.5

0.5

� = [0.166 y 4 y z] dy dz2

y 0.5

0.5

z 0.5

0.5

��

= 0.083y4y

3z dz2

3

z 0.5

0.5

0.5

0.5

��

��

�

��

� = 0.333 z dz

z 0.5

0.5

�� = [0.1666 z ]2

0.50.5� = 0

Example 2.10.8

Solution : D = 2 2 2� � ��z a a z� cos

i) There are three surfaces S1, S2 and S3 as shown in the Fig. 2.20

For S1, z = 1, � varies from 0 to 5 and � varies from 0 to 2�.



dS = � � �d d a z

For S2, z varies from –1 to 1 and � varies from 0 to 2�with � = 5.

d S = � � �d z d a

For S3, z = –1, � varies from 0 to 5 and � varies from 0to 2�.

dS = � �� d d �a z

D S5� d

S1

=� �

� � � �'

� � %

8

�

�cos2

0

2

1

d d

z ��

a a z� 5 � 0

=� �

�0

�

�

31 2

20

5

0

2�

��

�

��

��

��

��

cosd =

1253

12

� ��

��

��

sin 22 0

2

=125

62� � = 130.8996

D S5� d

S2

= 2 21

0

2

� �'

�

�

z dz d

z 1�� ... � � 5

= ( )5 23

23

1

1

02� �

�

��

�

��

��

z� � 209.4395

D S5� d

S 3

=� �

�� '

� � %

8

�

�cos2

0

2

1

d d

z � ��

= ��

��

�

��

� ��

��

�

�3

0

52

0

2

3cos d 130.8996

� D S5� d = 130.8996 209.4395 130.8996� � � 209.4395

ii) 3 5 D = � �1 1�

44�

��

4 �4�

44�D

D Dzz

� �

= ( ) � � � �12

102 2 2

�44�

��

44�

44

� �zz

� � cos = 4 2z

� 3 5� D dv

v

= 4 2

0

5

0

2

1

1

z d d dz

z

� � ��

�

��

= ( )43 2

3

1

1 2

0

5

02z�

��

�

��

�

��

�

��

�� = 4

23

252

2� � � �� 209.4395

��



z

� = 5

S1

S3

z = 1

z = –1

y

S2

x

Fig. 2.20


Example 3.3.4

Solution : The work done is given by,

W = � � �Q d

B

A

E L

Let us differential length dL in cartesian co-ordinate system is,

dL = dx dy dza a ax y z� �

� E L�d = � � � �� 8 xy 4x dx dy dz2a a a a a ax y z x y z

= � � �8 xy dx 4x dy dz2

As a a a a a ax x y y z z� � � 1, other dot products are zero.

� W = � � ��Q 8 xy dx 4x dy + dz

B

A2

= � � � �

��

��

� � �Q 8 xy dx x dy dz

B

A

B

A2

B

A

4

Case 1 : The path is y 3x z, z x 42 � �

� y = 3x x 42 � � differentiate

� dy = (6x 1) dx�

For

B

A

8 xy dx� � � The limits are x = 1 to x = 2

For

B

A

4 y� � x2 � The limits are y = 8 to y = 18

For

B

A

dz� � The limits are z = 5 to z = 6


3 Electric Work, Energy and Potential

(3 - 1)

� W = � � � �

�

��

�

�� Q 8 xy dx x dy dz

x= 1

2

y= 8

182

z= 5

6

4

Using y 3x x 42 � � and dy = (6x + 1) dx and changing limits of y from 8 to 18 interms of

x from 1 to 2 we get

� W = � � � ��

�

� � � �Q 8x 3x + x+ 4 dx x 6x+ 1 dx dz

x = 1

22

x = 1

22

z = 5

6

4�

�

��

= � � � ��

�� Q 24x 8x 32x dx 24 x 4x dx dz

x= 1

23 2

x= 1

23 2

z = 5

6��

�

��

= � ��

��

��

��

Q 6x83

x 16x 6x43

x z4 3 2 4 3

x 1

2

56

= � �� Q 256 1 6 255 1530 JCase 2 : Straight line path from B to A.

To obtain the equations of the straight line, any two of the following three equations of

planes passing through the line are sufficient,

B (1, 8, 5) and A (2, 18, 6)

(y y )B� = � �y y

x xx xA B

A BB

��

�

(z z )B� = � �z z

y yy yA B

A BB

��

�

(x x )B� = � �x x

z zz zA B

A BB

��

�

Using the co-ordinates of A and B,

y – 8 =18 82 1��

�(x 1)

� y – 8 = 10 ( x – 1)

� y = 10x – 2

� dy = 10 dx

And z – 5 =6 5

18 8��

�(y 8)

� z – 5 =110

(y 8)�

� 10 z = y + 42

Electromagnetic Theory and Transmission Lines 3 - 2 Electric Work, Energy and Potential


Now W = � � � �

�

��

�

�� Q 8 xy dx x dy dz

x= 1

2

y= 8

182

z= 5

6

4

= � � � � �

�

��

�

�� Q 8 x(10x 2) dx x 1 dx) dz

x= 1

2

x= 1

22

z= 5

6

4 0(�

= � ��

� �

��

��

��

��

!�

"�

Q3

xx x

3z3

2 3

x 1

80 162

402

56

= � �� Q 213.33 32 106.667 26.667 8 13.33 1

= � �� Q 6 255255 1530 JThis shows that irrespective of path selected, the work done in moving a charge from Bto A remains same.

Example 3.3.5

Solution : The line charge along the

z-axis and the circular path along which

charge is moving is shown in the

Fig. 3.1.

The circular path is in xy plane such that

its radius is r 1 and centered at the line

charge.

Consider cylindrical co-ordinate system

where line charge is along z-axis.

The charge is moving in a # direction.

� dL = r d# #aThe field E due to infinite line charge

along z axis is given in cylindrical

co-ordinates as,

E =$%&

L

02 ra r

The circular path indicates that dL has no component in a r and a z direction.

� dL = r d# #a

� W = � �� Q d Q2 r

r d

initial

final

0

2L

0E L a ar

%

#$%&

# = � �� Q2

d

0

2L

0

%

#$%&

# a a 0r

As a ar � # 0 as ' ()* between a ar and # .

This shows that the work done is zero while moving a charge such that path is always

perpendicular to the E direction.



y

r1

r1

d = r dL a# #

Infiniteline charge

Movement of charge Q inz=0 plane (xy plane)

Circular pathhaving radius r1

z

x

Fig. 3.1

Example 3.3.6

Solution : The charge is moved from B(2, 1, –1) to A (4, 2, –1).

Now, W = Q d

B

A

� +E L where d d d dx y zL a a dx y z � �

� E L+d = x a Lx +d = x dx … a a a ax y x z+ + = 0

� W = � �Q dx

x = 2

x = 4

x = � �

��

��

( 5)x2

2

2

4

= 5162

42

� ��

��

= 30 J

Example 3.3.7

Solution :

Note The paths are located at the points. Hence charge is moved through , L rather

than from one point to other. It is moved at a point in the direction , L through

distance , L. Hence the length is differential and work done will be also

differential. There is no need of integration.

� dW = – QE L�,= – Q � � � �6y z 12 xyz 6 xy2 2a a a a a ax y z x y z� � � � �� 3 5 2

= – Q� �� 18 y z 60 xyz 12 xy2 2 10 6 ... as -m

= � �� 2 10 106 618 y z 60 xyz 12 xy2 2

a) At � �P1 0 3 5, , substitute x = 0, y = 3, z = 5

� dW = � �� 2 10 81012 = 1620 pJ

b) At � �P2 1 1 0, , substitute x = 1, y = 1, z = 0

� dW = � �� 2 10 0 0 1212 24 pJ

c) At � �P 0.7, 2, 0.43 � � substitute x = – 0.7, y = – 2, z = 0.4

� dW = � �� 2 10 12 28.8 33.6 33.6 76.8 pJ



Example 3.3.8

Solution : The path along which the charge

is moved is shown in the Fig. 3.2

As path is straight line from B to A and the

line is in xy plane (z = 0 plane), the equation

of line can be easily obtained as,

y = mx ... Passing through origin

Now, B (4, 2, 0) and A (0, 0, 0)

� y =y y

x xx

0 20 4

x12

xA B

A B

��

��

� x = 2y i.e. dx = 2 dy

dL = dx dy dya a ax y z� � ... Cartesian

� W = � � �Q d

B

A

E L = � ��

��

��

�� Q

x2

2y 2x dx dy dz

B

A

a a a a ax y x y z

= � ��

�� Q

x2

2y dx (2x) dy

B

A

... a a a ax x y y� � 1

= � ��

��

��

��

!�

"�� Q

x2

2y dx+ 2x dy

x= 4

0

y= 2

0

... (1)

Using x = 2y and dx = 2dy and changing limits interms of y, in equation (1),

W = � ��

��

�

��

��

��

!�

"�� Q

2y

22y (2dy)+ 2y dy

y= 2

0

y= 2

0

2

= ��

��

!�

"� �� Q 6y dy+ 4y dy Q

y = 2

0

y = 2

0 6y

2

4y

2

2 2

2

0

�

��

��

= � � � � ��20 10 [ 12 8]=6 + 400 J-

Example 3.3.9

Solution : Q = 2 C, B(2, 0, 0), A(0, 2, 0), E = 12 x 4 ya ax y�

W = – Q

B

A

� �E dL where dL = dx dy dza a ax y z� �

= � � �� 2 12 x 4 y dx dy

(2,0,0)

(0, 2 , 0)

x y x( ) (a a a a y zdz� a )



y

x

z

4

2 B(4,2,0)

A(0,0,0)

Fig. 3.2

= � ��2 12 x dx 4 y dy

(2,0,0)

(0, 2 , 0)

… (a a a a a ax y x z y z� � � 0)

= �

��

��

�

��

��

212 x

2

4 y

2

2

x 2

x 0 2

y 0

y 2�

��

��

!�

"�= � � �2 32 = + 64 J

Example 3.3.10

Solution : W = � ��Q dE L = � � � � ��( 2) [y x ] [dx dy dz ]a a a a ax y x y z

= � �2 y dx x dy� �� …x = 2y 2

= � �2 y 4y dy 2y dy2� �� …dx = 4y dy

= � �2 4 y dy 2 y dy 2 6 y dy2 2

1

22� � ��

��

��

= � �12y

34 (2) (1) 4 (8 1)

3

1

2

3 3

��

��

� � 28 J

Example 3.3.11

Solution : The work done is given by,

W = � � �Q d

B

A

E L

Now dL = dr r d rsin da a ar � �' ' #' #

Consider the path as shown in the Fig. 3.3

Along path I, d' = d 0#

�E L�d = 5e drr 4� � ... (1)

Along path II, dr = d' = 0

�E L�d =10

rsinrsin d

'' #�

= 10 d# ... (2)

Along path III, dr = d# = 0

� E L�d = 0 ... (3)

� W = � � �

�� Q [5e dr 10 d 0]

0

2

0

4

r 0

2r 4

#

%

'

%

#



x

y

z

I

II

III

P(2, , )%4

%2

Fig. 3.3

= � ��

��

��

� ��

��

�

�5 10

5e1 4

10[ ] 06r 4

0

2

02# %

!�

"�= � �

�

��

��

��

��

!�

"�

��

5 105e

1 4

10

26

r 4

0

2%

= � � � � ��

!"

�� 5 10 20 e 20 e10

26 0.5 0 %

117.88 J-

Example 3.5.8

Solution : The potential of A due to point charge Q at the origin is given by,

VA =Q

4 r0 A%&and A (2, 2, 3), Q at (0, 0, 0)

where rA = ( ) ( ) ( )2 0 2 0 3 0 172 2 2� � � � �

� VA =0.4 10

4 8.854 10 17

9

12

�

� � �

�

�%0.8719 V ... The reference is at infinity.

Example 3.5.9

Solution : Now the Q is located at (2, 3, 3).

The potential at A is given by

VA =Q

4 R0 A%&where

RA = r r� .

= ( ) ( ) ( )2 2 2 3 3 3 12 2 2� � � � � ... by distance formula

� VA =0.4 10

4 8.854 10 1

9

12

�

� � �

�

�%3.595 V

Example 3.5.10

Solution : VAB = V VA B�where V VA Band are the absolute potentials of A and B.

Now VA = 3.595 V ... As calculated earlier.

VB =Q

4 R0 B%&where RB is distance between point B and Q (2, 3, 3)

� RB = ( ) ( ) ( )� � � � � � 2 2 3 3 3 3 42 2 2

� VB =0.4 10

4 8.854 10 4V

9

12

�

� � �

�

�%0 8987.

� VAB = V VA B� = 3.595 – 0.8987 = 2.6962 V



r'

rO

Q(2,3,3)

A(2,2,3)

R =|r–r'|A

Fig. 3.4

Example 3.5.11

Solution : Let Q C1 3 - , Q C2 �4 -

and Q C3 5 -

The potential of A due to Q 1 is,

VA1 =Q

4 R1

0 1%&

and R1 = ( ) ( ) ( )1 0 0 0 1 02 2 2� � � � �

= 2

� VA1 =3 10

4 2

6

0

�

�

�

%&= 19.0658 kV


VA2 =Q

4 R2

0 2%&

and R2 = � �1 2 0 1 1 3 62 2 2� � � � � � [ ( )] ( )

� VA2 =� �

� �

�4 10

4 614.6769 kV

6

0%&


VA3 =Q

4 R3

0 3%&

and R 3 = � �1 0 0 4 1 2 262 2 2� � � � � � ( ) [ ( )]

� VA3 =5

8 8132�

�

�10

4 26kV

6

0%&.

� VA = V V VA1 A2 A3� � � 13.2021 kV

Example 3.5.12

Solution : The charges are placed as shown in the Fig. 3.6.



A(1,0,1)

R1

R2

R3

Q (0,0,0)1

Q (2,–1,3)2

Q (0,4,–2)3

Fig. 3.5

QP Q Q Q

1 2 4 8

x

y

– ax

Fig. 3.6

The potential at a point due to a point charge is,

V =Q

4 R%&)

For charge at x = 1, � �R 1 210 , For charge at x = 2, � � � �R 2 22

1

For charge at x = 4, � �R 4 232 , For charge at x = 8, � �R 8 24

3

� VP =� � � � � � � �

V V V VQ

1 2 3 4� � � � �

� � � �

��

�... ....

104

1

2

1

2

1

2

1

2

9

0 1 2 3%&) �

��

From the series, / � / � / /r r r2 � � 0... the sum given by, S =/

1 r�

Comparing the series, / = 1, r =12

hence S =1

112

2�

� VP =Q

8.854 10 12

�

� ��

�

�

10

42

9

%= 17.9754 Q V

Similarly E =Q

4 R2%&)a R where a R = �a x

� EP =� � � � � �

� �Q��

�

��

�

��

�104

11

2

1

2

1

2

9

2 2 2 2 3%&)... a x

Comparing the series, / = 1, r =1

22hence S =

/1 r�

�

1

11

2

43

2

� EP =� � � �

Q

8.854 10 12

� �

� ��

�

�

10 4 3

4

9 /

%a x = – 11.983 Q a x V/m

Example 3.5.13

Solution : Q = 15 nC at origin

i) Find V1 at P(2, –3, –1)

� V1 =Q

r4 0 1%&and r1 = � � � � � �2 0 3 0 1 02 2 2– – – – –� � = 14

� V1 =15 10

4 8 854 10 14

9

12

�

� � �

–

–.%= 36.0311 V

ii) V = 0 at (6, 5, 4)

� V1 =Q

rC

4 0 1%&� and V1 = 0 at (6, 5, 4)



� 0 =Q

rC

4 0 2%&� where r2 = � � � � � �6 0 5 0 4 02 2 2– – –� � = 77

� 0 =15 10

4 8 854 10 77

9

12

�

� � �

–

–.%+ C i.e. C = –15.3637

� V1 =Q

r4 0 1%&– 15.3637 = 36.0311 – 15.3637 = 20.6673 V

Example 3.5.14

Solution : E = � � �6 y

x

6

x5

2a a ax y z

VAB = � +�B

A

E dL where dL a a ax y zdx dy dz � �

Now a a a a a ax x y y z z+ + + = 1 and all other dot products are zero.

� E L+d = � � �6 y

xdx

6x

dy 5 dz2

� VAB = � � � ��6y

xdx

6x

dy 5 dz

B

A

2

To obtain the integral as it does not depend on the path from B (4, 1, 2) to � �A 7, 2, 1� we

can divide the path as,

Path 1, B (4, 1, 2) to � ��7 1 2, , � only x varies, y = 1, z = 2.

Path 2, � ��7 1 2, , to � ��7 2 2, , � only y varies, x = – 7, z = 2.

Path 3, � ��7 2 2, , to A � ��7 2 1, , � only z varies, x = – 7, y = 2.

� VAB = ��

�

�

� �x 4

x 7

2y 1

y 26 y

xdx

6x

dy 5 dz

z 2

z 1

��

��

!�

"�

�

y = 1 x = – 7

= � � � �

�

� �6

1

xdx

67

dy 5

x 4

7

2y 1

2

z �

��

��

!�

"�2

1

dz

= � � � ��

��

� ��

��

!�

"

�6

1x

67

y 5 z4

7

12

21

�

= � � � ��

��

��

!"

617

14

67

2 1 5 1 2

= � �� 2.3571 0.85714 5 = + 8.2142 VAlternatively find the equations for straight line path from B to A by using,

y y B� = � �y y

x xx xA B

A BB

��

� and � �z zz z

y yy yB

A B

A BB�

��

�



and using the relations between x, y and z solve the integrals. From the above equationswe get, x = – 11 y + 15 and z = – y + 3 so use y interms of x for first integral and xinterms of y for the second integral and integrate.

Example 3.5.15

Solution : Potential due to the point charge,

V =Q

4 r0%&… r = Distance of point from Q

� VA =Q

4 r0 A%&and V

Q4 rB

0 B

%&

� VAB = V VQ

41

r1rA B

0 A B� �

��

��%&

= ��

� ��

��

�

�

20 10

4 8.854 10

10.5

10.1

10

12%= 143.8038 V ... Q is negative

Example 3.5.16

Solution : Q = 5 nC, V = 2 V at (0, 6, – 8), Q is at origin (0, 0, 0).

i) A (– 3, 2 , 6)

rA = – 3 a x + 2 a y + 6 a z , rA = 9 4 36� � = 7

� VA =Q

4 rA%&0+ C

VR = 2 V at (0, 6, – 8) hence rR = 6 a y – 8 a z , rR = 6 82 2� = 10

VR =Q

4 rR%&0+ C i.e. 2 =

5 104 10

9

0

��

�

%&+ C i.e. C = – 2.4938

� VA =5 104 7

9��

�

%&0– 2.4938 = 3.926 V

ii) B (1, 5, 7)

� rB = a x + 5 a y + 7 a z , rB = 1 25 49� � = 75

� VB =Q

4 r%&0 B+ C =

5 10

4 75

9

0

�

�

�

%&– 2.4938 = 2.6952 V

iii) VAB = VA – VB = 3.926 – 2.6952 = + 1.23 V



Example 3.5.17

Solution : The charges are shown in the

Fig. 3.7.

The distances between the charges and

point P are,

R1 = � � � �0 1 0 z 02 2 2� � � � = 1 z2�

R2 = � � � � � �0 0 1 z 02 2 2� � � � = 1 z2�

R 3 = � � � � � �0 1 0 z 02 2 2� � � � = 1 z2�

R4 = � � � � � �0 0 1 z 0 1 z2 2 2 2� � � � �

The potential at P due to all charges is,

a) VP = V V V V1 2 3 4� � �

= � �14 %&) R

Q Q Q Q1 2 3 4� � �

where R = R R R R 1 z1 2 3 42 �

� VP =4 6 10

4

9� �

� � � �

�

�

�% 8.854 10 1 z12 2

215.7058

1 z 2V

b) To find � �VP max , find dV/dz = 0

� � �ddz

215.7058 1 z2 0.5�

��

��

= 215.7058 (– 0.5) � �1 z2 1.5�

�(2z) = 0

Thus z = 0 or 1 z2� = 0 i.e. z = 1 j but z coordinate can not be imaginary.

� At z = 0, VP is maximum i.e. � �VP max = 215.7058 V

c) NowdVdz

= � �� 215.7058 0.5 1 z 2 z2 1.5� �

�=

� �215.7058 z

1 z 2 1.5�

d) To find its maximum value,d

dzdVdz

��

!"

= 0

� � �ddz

215.7058 z 1 z2 1.5��

��

!"

�= 0

� 215.7058 � �� z 1.5 1 z 2 z 1 z2 2.5 2 1.5� � � ��

��

!"

� �= 0



O Q

B(0,1,0)2

C (–1,0,0)Q

D (0,–1,0)4

A(1,0,0)Q1

R4

R1R2

R3

P(0,0,z)

z

Q3

Fig. 3.7

� 215.7058 � �1 z3 z

1 z12 1.5 2

2� �

��

��

��

!�

"�

�= 0 i.e.

�

��

3 z

1 z1

2

2= 0

� � � �3 z 1 z2 2 = 0 i.e. z2 =12

= 0.5

� z = 1 0.5 = 1 0.7071 for � �dVdz

max

� � �dVdz

max =� �

� �� 215.7058 0.7071

1 0.7071 2 1.5�

= 83.024 V/m

Example 3.6.5

Solution : $L = 1 nC/m

along x-axis and find V at

A (0, 1000) mm.

Consider elementary

charge dQ on length dx at

distance x from origin.

� dQ = $L dx

Distance of point A from

charge dQ is,

R = x 12 �

� d VA =dQ

4 R0%&

=$

%&

L

02

dx

4 x 1�

� VA =

�

� �

� ��

0.5

0.5L

02

0

0.5dx

4 x 12

$

%&

$L

02

dx

4 x 1%& �... changing limits

=2

4dx

x 1

L

0 0

0.5

2

$%& �

�.... �

� � �

��

dx

x ax x a

2 2

2 2ln

=2

4x x 1L

0

2

0

0.5$%&

ln � ��

��

��

!"

= � �2 1 10

4 8.854 100.5 0.5 1

9

122� �

� ��

��

��

�%ln ln 1

� !"

� VA = 8.649 V



x

x

y

z

0 dx

C

(+0.5,0)in m

2

1m

A(0,1)m

(–0.5,0)in m

B

V = ?A

R = x +123

L

Fig. 3.8

Example 3.6.6

Solution : Q = 10 8� C, r = 5 m, h = 5 m. The

ring is shown in the Fig. 3.9.

$L =Q

circumference=

102 r

8�

%

=1010

8�

%

= 0.3183 nC m

Consider the differential length dL on the ring.

� dQ = $L dL

And dL = r d 5 d# # ... in xy plane

dQ = 5 dL$ #

� dV =dQ

4 R0%&and R = r z2 2�

� dV =5 d

4 5 5

L

02 2

$ #

%& �=

5 d

4 5 5

L

02 2

$ #

%& �=

5 d

4 50L

0

$ #

%&... r = z = 5

� V =

#

%%$ #

%&

$

%&#

0

2L

0

L

0025 d

4 50

5

4 50[ ]

� � =

5 0.3183 10 2

4 8.854 10 50

9

12

� � �

� � �

�

�

%

%12.7102 V

Example 3.7.3

Solution : Note that the charge is distributed over the entire surface of disc so it is a

surface charge. The surface charge density is,

$ S =Total charge

Area r (1.061 nC

2 2

�

403

403

2% % )/m2

V =$&s

02a h h2 2� �

��

with a = 2 m, h = 5 m

� V = 23.0776 V

Example 3.7.4

Solution : The ring lies in z = 0 i.e. xy plane as shown in the Fig. 3.10.

Consider the differential surface area dS at point P at a distance of r from the origin.

Hence differential charge dQ is

dQ = $S dS



z

P(0,0,5)

z

R

r = 5 m

O

a#dL

$L

y

x

Fig. 3.9

The dS in the xy plane is r dr d#

� dQ = $S r dr d#

� dV =dQ

r

r dr d

4 rS

4 %&$ #

%&) )

� V =

#)

4%

)

$ #%&� �

�

r R

R 1S dr d

4

= � � � �$%&

#)

%SRR 1

4r �

02

= � �� $%&

%)

S

4R 1 R� � 2

=$&S

02V

This shows that the potential at the origin due to the ring is independent of the inner

radius R.

Example 3.7.5

Solution : The charge is distributed along a ring so it is a line charge. Let .r = Radius of

ring = 5 m.

$L =Total charge

Circumference=

10

2

8�

.% r=

1010

8�

%= 3.183 × 10– 10 C/m

The ring is shown in the Fig. 3.11.

Consider the differential length d .L on the

ring at point P.

dQ = $L d .L

But d .L = .r d# = 5 d#

� dL. = 5 d# a #

� dQ = 3.183 × 10– 10 × 5 d#

dVA =dQ

4 R%&0

where R = Distance between A and

P = z (r )2 2� . = 50

� dVA =3.183 10

4 50

10� �

�

� 5

0

d#

%&

= 2.0228 d#



P

dSR+1

r

O

R

x

z

y

$S

Fig. 3.10

dL'

r' = 5 m

R = z +(r')32 2

$L

P

O

Ring inz = 0 plane

z = 5

A(0,0,5)

y

x

z

a#

Fig. 3.11

� VA = 2.0228 d##

%

�

0

2

= 2.0228 [# %]

02

= 2 % × 2.0228 = 12.7101 V

Now the same charge is distributed over a disc of .r = 5 m

� $S

=Total charge

Area=

10 8

2

�

.%( )r=

1025

8�

�%= 1.2732 × 10– 10 C/m2

Let the disc is placed in x-y plane as shownin the Fig. 3.12 with z-axis as its axis.

Consider differential surface d .S at point Phaving radial distance .r from the origin.

d .S = .r d .r d#

dQ = $S

d .S = $S

.r d .r d#

dVA =dQ

4 R%&0

=$ #

%&

S

r z

. .

. �

r dr d

4 (02 2)

…R = Distance AP

� VA = dVA� =$ #

%&#

%S

r r

. .

. �.��

r dr d

4 (

5

02

00

2

25)…z = 5 m

Put ( .r ) 2 + 25 = u2 i.e. 2 .r d .r = 2u du

For .r = 0, u1 = 5 and .r = 5, u2 = 50

� VA =$ #

%&#

%S

uu

u du d

4

u 2

010

2

��

=$%&

S

4 0[u] u 1

u 2[ ]# %

02

=1.2732 10

4 8.854 10[ 50 5]

10

12

�

� ��

�

�%[2% – 0] = 14.8909 V

Example 3.8.1

Solution : Consider a sphere of radius R with a uniform charge density $v .

Case 1 : Let point P is outside sphere (r > R).

The E is directed radially outwards, along a r direction.



dS'

r'

$S

P

z

A(0,0,5)

y

x

z

R

O

Fig. 3.12

dS = r sin d d2r' ' # a ... Normal to a r

� d5 = D S�d ... Gauss’s law

= D r sin d dr2a r � ' ' #

= D r sin d dr2 ' ' # (a ar r� 1)

� 5 =

S� � ��

D Sd D r sin d d Q

0

2

0

r2

#

%

'

%

' ' #

Solving, Dr =Q

4 r 2%i.e. D =

Q

4 r 2 r%

a

and E =Q

4 r02 r

%&a ... For r > R

Now V = �� E Ld and d dr rL a

� V = � �0 0� ��

r

r

02 r r

r

r

02

Q

4 rdr

Q

4 rdr

%& %&a a

Key Point The limits to be taken against the direction of the E i.e. from r = 0 to r.

� V = � � ��

��

�0 0�

Q4

1

rdr

Q4

1r

K0 r

r

2 0 r

r

%& %&= � �

Q4 r

K0%&

At r = 0, V = 0 i.e. K = 0 hence V =Q

4 r0%&

Now total charge contained by sphere is,

Q = Volume of sphere � $ % $v3

v43

R

� V =

43

R

4 rR3r

3v

0

3v

0

% $

%&$&

... r > R ... (1)

This is potential outside the spherical shell.

Case 2 : Let point P is inside sphere (r < R).

D = Dr ra and

dS = r sin d d2r' ' #a

� d5 = D S� d D r sin d dr2 ' ' #

� 5 = Q d

S

� �D S



P

Gaussiansurface

E

++

+ ++ +

+

+++

+

+

+++++

+

+ +

++ +

+

+++ +

dS

Chargedsphere

R

Fig. 3.13

P

Gaussiansurface

Chargedsphere

R

r

D a= Dr r

E a= Er r

Fig. 3.14

=

#

%

'

%

' ' # � �

0

2

0

r2D r sin d d = 4 r D2

r%

� Dr =Q

4 r 2%and D =

Q

4 r 2 r%

a

The charge Q enclosed by radius r < R must be considered.

� Q =

v

v

0

2

0 r 0

r2dv r sin dr d d� � � �

$ ' ' ##

%

'

%

=43

r 3v% $

� D =

43

r

4 r

r3

3v

2 rv

r

% $

%

$a a i.e. E =

Da

&$&0

v

0r

r3

� V = �� E Ld and d dr rL a

The limits of r are from r = R to r against the direction of E.

� V = � �

��

��

��

r R

rv

0

v

0

2

R

r

1r3

dr3

r2

K$&

$&

= � ��

6r R Kv

0

2 21

$&

... (2)

From equation (1), for r = R, V =R3

2v

0

$&

�R3

2v

0

$&

= 0 K1� ... (3)

Using in (2),

V = � ��

�

��

��6

r RR3 3

R r2

Rv

0

2 22

v

0

v

0

2 22$

&$&

$&

= � �$

&v

2 2

0

3R r

6

�... r < R ... (4)

Example 3.9.4

Solution : a) The charges are shown in the Fig. 3.15 (a).

For z > 1.5 m,

E1 =$&S1

2 0a z due to $ S1

E2 =$&S2

2 0a z due to $ S2

� E = � �12 0&

$ $S1 S2� a z = � �12

50 500

0 0&& &� a z



= 0 V/m for z > 1.5 m

For – 0.5 < z < 1.5,

E1 = � �$&S1

2 0�a z and

E2 = � �$&S2

2 0a z

� E = � �12 0&

$ $� �S1 S2 a z

= �50a z V/m

For z < – 0.5,

E1 = � �$&S1

2 0�a z and

� �E a2 z �$&S2

2 0

�E = � �12 0&

$ $� �S1 S2 a z

= � �12

50 500

0 0&& &� � a z

= 0 V/m

� E = 0 V/m for z > 1.5 m

= � 50 a z V/m for – 0.5 < z < 1.5

= 0 V/m for z < 1.5

b) To find potential at any point on z axis.

dL = dx a a ax y z� �dy dz

Select any point A (x, y, z) and B (x, y, – 0.5)

� VAB = � �� E L a a a az x y zd dx dy dz50

= � �

�50

z 0.5

z z

dz ... z varies from – 0.5 to z for points B to A.

= � � � �� 50 z 50 z 0.50.5z V

But VAB = V VA B�

where VA is potential at any point z



$S2

$S1

E1

E1

E1

E2

E2

E2

O

z

z = 1.5 m

z = –0.5 m

x

y

Fig. 3.15 (a)

VB is potential at z = – 0.5 which is zero.

� VA = 50 [z + 0.5] V

= V(z)

The graph of V(z) as a function of z is

shown in the Fig. 3.15 (b).

For z > 1.5 and z < – 0.5, the potential is

zero as E is zero in that region.

Example 3.9.5

Solution : The various charges are shown in the Fig. 3.16.

There are three charge configurations.

Case 1 : Point charge Q 1 200 %&) C at C � �3 1 2, ,� � .

VP =Q

RC1

114 %&)

� where C1 = Constant

R1 = � � � �� 5 3 6 1 7 22 2 2� � � � � �

� R1 = 78 ... Distance between P and C

To find C1, V = 0 V at Q (0, 0, 1)

� VQ =Q

RC1

214 %&)

�

where R2 = � � � �� 0 3 0 1 1 22 2 2� � � � � � = 11

� 0 =200

C1%&

%&)

)4 11� i.e. C1 = – 15.0755



–0.5 1.5

z

100 V

V(z)

Fig. 3.15 (b)

P(5,6,7) x = –3plane

dS$S

yQ

C(3,–1,2)1

$L

x

z

Fig. 3.16

� VP =200 %&

%&)

)4 78�– 15.0755 = – 9.4141 V

Case 2 : Due to line charge along x axis.

VPQ =$%&)

L Q

P

r

r2ln ... Potential difference

As line charge is along x axis, any point on it (x, 0, 0).

� rQ = � � � �0 0 1 02 2� � � = 1 ... 6 Distance from line charge

and rP = � � � �6 0 7 0 852 2� � � ... x not considered

� VPQ =40 %&%&

)

)21

85ln = – 44.4265 V

But VPQ = V VP Q� and V 0 VQ

� VP = V VPQ Q� = – 44.4265 V ... Absolute potential of P

Case 3 : Surface charge in the plane x = – 3 i.e. parallel to yz plane.

Note : As E due to infinite surface charge is known use V dABB

A

� � �E L.

So E =$&S

2 0a x ... a x is normal to yz plane

Point P is infront of plane as x co-ordinate of P is 5 hence � a x .

dL = dx a x + dy a y + dz a z

� E L�d =$&S

2 0dx ... a a a ax y x z� � = 0

� VPQ = � �Q

PS$&2 0

dx ... Potential between P and Q

xQ = 0 and xP = 5 hence

VPQ = ��0

5

02

$&S dx = � � �

52

52

8

0

0$&

&&)

S = – 20 V

But VPQ = V VP Q� and V 0 VQ

� VP = – 20 V ... Absolute potential of P

� Total VP = – 9.4141 – 44.4265 – 20 = – 73.8406 V

Example 3.9.6

Solution : The two line charges are shown in the Fig. 3.17.

Now V = 100 V at the origin O (0, 0, 0).

Let us obtain potential difference VPO using standard result.



Case 1 : Line charge 1

� VPO1 = �

��

��$%&

L

0

O1

P12

r

rln

where r and rO1 P1 are perpendicular

distances of points O and P from the line 1.

The line 1 is parallel to y-axis so do not use

y co-ordinates to find r and rO1 P1.

� rO1 = ( ) ( )1 0 2 0 52 2� � �

� rP1 = ( ) ( )1 4 2 3 102 2� � �

� VPO1 = �

��

��

$%&

L

0249.8386ln

5

10

But VPO1 = V VP1 O�

where V 100 VO

� – 49.8386 = V 100P1 �

� VP1 = 50.16 V ... Absolute potential of P due to line charge 1

Case 2 : Line charge 2, which is parallel to z-axis.

Do not consider z co-ordinate to find perpendicular distance.

� rO2 = ( ) ( )� � � � 1 0 2 0 52 2

and rP2 = ( ) ( )� � � � 1 4 2 1 262 2

� VPO2 =$%&

L

02118.5417 Vln

5

26

��

��

But VPO2 = V VP2 O� where VO = 100 V

� VP2 = – 118.5417 + 100 = – 18.5417 V

This is absolute potential of P due to line charge 2

� VP = V VP1 P2� = 50.16 – 18.5417 = 31.6183 V

Note Students can use the method of using consant C to find absolute potential of P due

to line charge 1 and line charge 2. Adding the two, potential of P can be obtained. The

answer remains same. For reference, the constant C C1 2 = 215.721 for both the line

charges.

Example 3.12.7

Solution : E = �7 V = �88

�88

�88

��

��V

x

V

y

V

za a ax y z



P

y

x = –1, y = 2

B(–1,2,z)

x

z

O(0,0,0)

Line 1

Line 2

A(1,y,2)

x = 1, z = 2

Fig. 3.17

88

V

x= � �

� �

� � � �2 y 2x 0 4

2x

x y

4xy8x

x y2 2 2 2 2 2� �

�

�

�

��

�

��

�

88

V

y=

� � � �2x 0 4

2y

x y

2x8y

x y

2

2 2 22

2 2 2� �

�

�

�

��

�

��

�

88

V

z= 0 20 0 20� �

� E =

� � � ��

�

�

��

�

��

� ��

�

��

�

��

4xy8x

x y

2x8y

x y2 2 22

2 2 2a ax y �

�

��

��

!�

"�

20a z

� E at P = � � � �� 60 0.0268 72 0.0112a a ax y z20

= + 59.9732 a x � 71.9888 a ay z�20 V/m

� D at P = E at P � &0 = 0.531 a x � 0.6373 a y � 0.177 a z nC/m2

Now $v = 7�D

and D = &0 E hence � �$ &v 7�E 0

7�E =88

�8

8�88

E

x

E

y

E

zx y z

=

� � � ��

88

��

�

��

�

��

88

��

�

��

�

�x

4xy8x

x yy

2x8y

x y2 2 22

2 2 2� ��

��

88 z

20

=

� � � ��

� ��

� � �

�

�

�

��

�

��

!

��

"

��

4y

x y 8 8x 2 x y

2x

x y

2 2 2 2 2

2 2 4

� � � � � ��

�

�

��

�

��

� �

� � �

�0

x y 8 8y 2 x y

2y

x

2 2 2 2 2

2� �y 2 4

�

�

��

�

��

!

��

"

��

�

��

�

��

– 0

=

� � � � � � � ��

��

��

��

�4y

8

x y

32 x

x y

8

x y

32 y

x y2 2 2

2

2 2 3 2 2 2

2

2 2 3

At P, x = 6, y = – 2.5 and z = 3.

� 7�E = 10 4.4816 10 0.01527 4.4816 10 2.651 103 3 3� � � � � �� = 10.00895



� $v at P = � �&0 7 � �� E 8.854 10 10.0089512 = 88.6193 pC/m3

Example 3.12.8

Solution : i) E = �7 �88

�88

�88

VVr

1r

V 1r sin

Va ar # '' # #a

��

= � � � �20

rsin cos

1r

cos10

rcos

1r s3 2

' # # ' 'a ar � �in

10

rsin sin

2'' # #�

��

��a

=20

rsin cos

10

rcos cos

10

rsin

3 3 3' # ' # 'a ar � � # #a

D = & & ' # ' # '0 0 3 3

20

rsin cos

10

rcos cos

10E a ar � �

rsin

3# #a

��

��

At point 22

0, ,%�

��

��, r = 2, '

%

2and # = 0

� D = 2.5 &012 222.135 10 C/ ma ar r � �

ii) W = � ��Q d

initial

final

E L

Now

initial

final

d� �E L = Potential difference between initial and final

Initial point � �A 1, 30 , 120* * and final point � �B 4, 90 , 60* *

�A

B

d� +E L = V V VAB A B �

VB =� �

� � � �V10

rsin cos

10

4sin 90 cos 60at B 2

at B2

* *' # = 0.3125 V

VA =� �

� � � �V10

rsin cos

10

1sin 30 cos 120at A 2

at A2

* *' # � 2.5 V

� VAB = V V 2.5 0.3125 2.8125 VA B� � � �

� W = � � � � � �Q V 10 10 2.8125 JAB6 28.125 - ... Same as above

Example 3.12.9

Solution : a) V = E ey

0x� �

��

�

��sin

%4

� E = �7 V = �88

�88

�88

��

��V

x

V

y

V

za a ax y z



88

V

x= � �E

ye0

xsin%4

1�

��

�

�� ... y is constant

88

V

y= E e

y0

x� �

��

�

��cos

% %4 4

... x is constant

88

V

z= 0 ... z is absent

� E = � � �

��

�� E e

yE e

y

40x

0xsin cos

% %9

%4

a ax y V/m

At P ( , , )0 1 1 , E = E 0.7071 0.5550 [ ]a ax y� V/m

b) V = E r cos0 '

� E = �7 V = �88

�88

�88

��

��V

r1r

V

r

Va a ar ' ' #' #

1sin

88

V

r= E0 cos ',

88

V

'= �E r0 sin ',

88

V

#= 0

� E = � �E E0 0cos sin' ' 'a ar V/m

Convert P � �0 1 1, , to spherical co-ordinates.

r = x y z 22 2 2� � , # = tan � 1 y

x%4

, ' cos� *1 zr

45

� E = � � �E [ 0.7071 a 0.7071 a0 r ' ] V/m

Example 3.12.10

Solution : i) E = – 7 V = –88

88

88

Vx

Vy

Vz

a a ax y z� �

��

��

V = 10 y x3 + 50 y V

� E = – [30 y x2 a x + (10x3 + 50) a y + 0 a z ]

At y = 0, E = – (10 x3 + 50) a y V/m

ii) E is only in a y direction, E = – (10 x3 + 50) a y

� E = – 7 V = –88

88

88

Vx

Vy

Vz

a a ax y z� �

��

��

Equating the two sides, – (10x3 + 50) = –88Vy

Integrating with respect to y, V = 10 x3y + 50 y + C

At y = 0, V = C = Constant



Thus as potential is constant along the surface y = 0, it is equipotential surface.

iii) dS = dx dz a y , E = – (10x3 + 50) a y

D = &0 E = – &0 (10 x3 + 50) a y

� Q = D S�� d

S

=

z

x

= 0 = 0

1 2

0310 50� � � �

x

& ( ) dx dz … a y � a y = 1

= – &010 x

450 x [z]

4

0

2

01�

��

��

= – 1.2395 nC

Example 3.12.11

Solution : V =100

z 1cos

2 �$ #, $ = 3, # = 60º, z = 2

i) V =100

2 1cos 60º=

2 �� 3 30 V

ii) E = –7 V = –V 1 V V

z88$ $

88#

88$ #a a a z� �

��

= � �–cos100 1 100

100#

$$ # $ #$ #

z 1 z 1–sin cos

–2 z

z2 2 2�

� ��

� ��

a a

� �1 2

�

��

�

��

�

��

��

!�

"�

a z � (1)

= –10 a 17.32 a 24 a z$ #� � V/m � Using given values

iii) E = 10 17.32 242 2 2� � = 31.24 V/m.

iv)dVdN

= E = 31.24 V/m

v) a N = –E

E0.32 a – 0.55 a – 0.768 a z $ #

vi) $v = � � � �7 7 7+ + +D E E& &0 0 = � �&$

88$

$ $$

8

8#88

#0

1E

1 E Ezz� �

��

��

From (1), E$ =–100 cos

z 12

#

�, E# =

100 sin

z 12

#

�, Ez =

� �200 z cos

z 12

$ #

�2

� $v = &$

88$

$ #$

88#

#0 2

100 100

1

1 cos

z + 1

12

– sin

��

��

�

��

��

z � ��

�

�

��

�

��

�

��

��

!�

"�

88

$ #z

200 z cos

z2 12



=� � � � � ��

&$

#$

#$ #0 2

100 100

1200

1 cos

z + 1

1 z 1 1 – z 2 z

2

2 2

� ��

��– cos

cosz

� ��

� �

2

2 4

1 2z

z 1

�

�

�

��

�

��

�

��

��

!�

"�

=� �

� �&

$ #&0 3 00

200�

�

��

��

!�

"�

� � �cos 1 3z

z + 1

200 3 cos 60º2

2

– � �–11

5 3

��

!"

= –233.75 $C/m3

Example 3.12.12

Solution : V = 2x2y – 5z, P(– 4, 3, 6)

� V = 2 4 3 5 62� � � � �( ) = 66 V

E = �7V = � � �

��

��88

88

88

Vx

Vy

Vz

a a ax y z = – [4xy 2x 5 ]2a a ax y z� � …(1)

� At P, E = � � �48 a 32 a 5 a V / mx y z

D at P = E at P � &0 = 0.425 a – 0.2833 a 0.0442 ax y z� nC/m2

� $v = 7 +D =88

8

888

Dx

D

yDz

x y z� � …(Use equation 1)

= &0 [ 4y 0 0]� � � = �4 0& y C/m3

� $v at P, $v = – 4 � 3 &0 = – 12 &0 = – 0.1062 nC/m3

Example 3.12.13

Solution : Refer example 3.12.12 for the procedure and verify the answers :

i) V= 251 V ii) E = 24a 30a 96ax y z� � V/miii) D E0 & = 0.215 a 0.2656 a 0.85 a nC/ mx y z

2� �

Example 3.12.14

Solution : Given V is in cylindrical system.

i) E = �7 V = –88

�88

�88

��

��V

r1r

V V

za a ar z# #

88

V

r= cos 2 2# # �

88

��

��

��

��

��

��r

1r

1

r 2cos

88

V

#= � � � �1

r1r

88

��

��

## #cos sin2 2 2 and

88

V

z= 0

� E = � � ��

��

��

cos sin2 2 2# ##

r

1r r2

a ar =cos sin2 2 2# #

#r r2 2

a ar �

At B � �2 30 1, ,* , E = 0.125 a r + 0.433 a # V/m



ii) $v = � � � �7 7 7� � �D E E& &0 0

7�E = � �1r r

r E1r

E E

zrz8

8�

8

8�88

#

#

=1r r

r

r

1r r2 2

88

��

��

88

��

��

cos sin2 2 2##

#+ 0 = � �1

r1

r

1r

2

r2 2cos cos2 2 2# � #

�

��

�

��

� $v = � �& &# # #

&0 0 02 4 2 3 2

7 � �

��

��

��

��E

cos cos cos

r r r3 3 3

At A (0.5, 60*, 1), $v =� �

� �

3 2 60120

0� � *

�cos &

&0.5 3

= – 0.106 nC/m3

Example 3.12.15

Solution : E = – 7V = –88

88

88

Vx

Vy

Vz

a a ax y z� �

��

��

88Vx

=88x

[3x2y + 2yz2 + 3xyz] = 6xy + 0 + 3yz

88Vy

=88y

[3x2y + 2yz2 + 3xyz] = 3x2 + 2z2 + 3xz

88Vz

=88z

[3x2y + 2yz2 + 3xyz] = 4yz + 3xy

� E = – {[6xy + 3yz] a x + [3x2 + 2z2 + 3xz] a y + [4yz + 3xy] a z }

At point (1, 2, – 1), x = 1, y = 2, z = – 1

� E = – 6 a x – 2 a x a y + 2 a z V/m

Example 3.13.7

Solution : The charges existing at the corners of an

equilateral triangle are shown in the Fig. 3.18

When Q 1 is placed, no other charge is present,

� WE1 = 0 J

When Q 2 is placed, Q 1 is present.

� WE2 = Q V QQ

4 R2 21 21

21

��

��%&)

= 21

4 12

4J

0 0�

�

��

��

%& %&

When Q 3 is placed, Q 1 and Q 2 both are present,

� WE3 = Q V + Q V QQ

4 RQ

Q

43 31 3 32 31

313

2

0

��

��

%& %&) R 32

��

��

� WE3 = 31

43

24

J0 0 0

��

��

��

�

��

��

%& %& %&1 19

4



1 m

1 m

1 m

Q = 1 C1

Q2Q32 C3 C

Fig. 3.18

� WE = W W WE1 E2 E30 0

� � � �02

49

4%& %&

=11

4 8.854 10 12% � � �

�9.8865 10 J10

Example 3.13.8

Solution : E = �7 V = �88

�88

�88

��

��Vx

Vy

Vz

a a ax y z

88Vx

= 1 + y,88Vy

= – 1 + x,88Vz

= 2

� E = � � � �� 1 y x 1a a ax y z2

At (1, 2, 3), E = � � � �� 1 2 1 1 2a a ax y z = � �3 2a ax z V/m

Now WE =12 0

2

vol� & E dv

E = � � � � � �1 y x 1 22 2 2� � � �

� | E|2 = � � � � � �1 y x 1 2 y 2y x 2x 62 2 2 2 2� � � � � � � �

WE =12 0

2

vol� & E dv and dv = dx dy dz

The cube is centered at the origin.

Thus x varies from – 1 to + 1, y from – 1 to + 1 and z from – 1 to + 1.

� WE = � �&0

2z 1

1

y 1

1

x 1

12 2y 2y x 2x 6

� � �� dx dy dz

Changing limits from – 1 to 1 to 0 to 1 of each integral, making it twice.

� WE = � � � �&0

22 2 2� � � � � �

� � �

z 0

1

y 0

1

x 0

12 2y 2y x 2x 6 dx dy dz

= 4 0&z 0

1

y 0

12

32xy 2xy

x3

x 6x

� � � � � �

��

��

dy dz

= 4 0

0

1

& xzy

3zxy

x3

yz x yz 6 xyz3

23

2� � � �

��

��

= 0.2361 nJ

Example 3.13.9

Solution : i) V =50r

� E = �7V = �88

�88

�88

��

��Vr

1r

Vr

Va a ar ' ' #' #

1sin



�88Vr

= �50

r 2,

88

88

V V' #

= 0

� E =50

r 2a r

� D = &&

0050

E a rr 2

� $v = 7 +D = � �1

r rr D

22

r88

� �0 0

= � �1

r rr

r

1

r r22

2 2

88

�

��

��

88

50500

0&

& = 0 C/ m3 ... Proved

ii) WE =12 0

2

vol� & E dv

E2

=� �50

r

2

4and dv = r dr d d2 sin ' ' #

� WE =&

' ' ## )

4%

'

%0

02 � � �

r a

b 2

4250

rr dr d dsin

= � � � �2500 0

0 02&

4' #% %� �

��

cos1r a

b

= 1250 � �� & %0 2 2 � � ��

��

��

��1b

1a

= 1.39 101a

1b

7� ��

��

� J

Example 3.13.10

Solution : E =106

r 6a r , � E

10

r

6

6, E

2

10

r

12

12

� WE =12

2

vol� & E dv =

12

vol

12

12

10

r� & dv

dv = r dr d# dz in cylindrical system

z = 0 to 200 mm i.e. 0 to 0.2 m, # = 0 to 2 % and r = 0.05 to 0.1 m.

� WE =12

1012&# )

4%

z 0

0.2

r 0.05

0.1

12

1

r � � � r dr d# dz

� WE = � � � � � �12

10 10012

02& # %

z1

r00.2

r 0.05

0.1

11� dr ... & :) &)

=10

2

13

0& � 0.2 � ��

��

��

�2 %

r10

10

0.05

0.1



= 55.6313� � � �

� � �

��

��

1

10 0.1

1

10 0.0510 10= 5.691 � 1013 J

Example 3.13.11

Solution : V = r z sin2 # ... cylindrical system

� E = �7 �88

�88

�88

��

��

VVr

Vr

Vz

a a ar z# #

= � � ��

2 r z sin1r

r z cos r sin2 2# # ##a a ar z��

WE =12

| | dv

vol0

2� & E

| |E = 4 r z sin r z cos r sin2 2 2 2 2 2 4 2# # #� �

� WE = � �&# # #0

vol

2 2 2 2 2 2 4 22

4 r z sin r z cos r sin� � � dv

dv = r dr d# dz

� WE =&

##

%0

z 2

2

0

/ 3

r 1

43 2 2 2

2r 4 z sin z

� � � � �� cos r sin dr d dz2 2 2# # #�

=&

##

%0

z 2

2

0

/ 32 2

4

1

2

24 z sin

r4

� � �

��

��

� z cosr4

r6

sin2 24

1

4 6

1

42# #

��

��

�

��

��

��

��

!�

"�d dz#

=&

##

%0

z 2

2

0

/ 32 2 2

2255 z sin 63.75 z cos

� � � �� 2 2682.5 sin d dz# # #�

=&

##

%0

0

/ 32

3

2

2 3

2255 sin

z3

63.75z3

��

��

��

� � �

��

��

��

��

!�

"��

2

22 2

22cos 682.5 sin z d# # #

= � �&# # # #

#

%0

0

/ 32 2 2

21360 sin cos sin d

� � �340 2730

= � �&# # #

#

%0

0

/ 32 2

24090 sin cos d

� � 340

=& #

##

##

%

#

%0

0

/ 3

0

/ 3

24090

1 cos 2d

1+ cos 2d

� �

��

2340

2

��

��

!�

"�

=&

##

##

#

%

#

0

0

3

02

40902

2

23402

2

2�

��

��

��

��

sin sin/ %/ 3��

��

!�

"�



=& % %0

22045

30.433

3402 3

0.433��

��

��

��

!"

= � � � �� &0

22045 0.6141 170 1.48019� � �

= 6.6735 nJ

Example 3.13.12


Fig. 3.19.

When Q1 is positioned, no other charge is present

hence W1 = 0 J.

When Q2 is placed, Q1 is present.

� W2 = Q2 V2, 1 =Q Q

4 R2 1

21% &0=

Q Q

4 d2 1

% &0

=2 10 2 10

4

9 9� � �

� � �

� �

�% 8.854 10 0.512= 71.9019 nJ

When Q3 is placed, both Q1 and Q2 are present.

� W3 = Q3 V3, 1 + Q3 V3, 2 = Q3Q

4 R

Q

4 R1

31

2

23% & % &0 0�

��

��

= 2 � 10– 9 2 104 1

2 104

9��

��

�

��

��

� �

% & % &0

9

0 05.= 107.8528 nJ

� WE = W1 + W2 + W3 = 179.755 nJ

Example 3.14.4

Solution : p = 3 a x – 2 a y + a z nCm at origin (0, 0, 0).

i) P(2, 3, 4)

� r = 2 a x + 3 a y + 4 a z , |r| = 29, a r =rr

V =p a r+

4 r 2% &0

=p r+

4 r 3% &0

=[( ) ( ) ( )]

( )

3 2 2 3 1 4 10

4 29

9

3

� � � � � � �

� � �

�

�% 8.854 10 12= 0.2302 V

ii) r = 2.5, ' = 30º, # = 40º

Converting given spherical to cartesian co-ordinates

x = r sin ' cos # = 0.9575, y = r sin ' sin # = 0.8034, z = r cos ' = 2.165

� r = 0.9575 a x + 0.8034 a y + 2.165 a z , |r| = 2.5

� V =p a r+

4 r 2% &0

=[(3 0.9575) ( 2 0.8034) (1 2.165)] 10

4 8.854 1

9� � � � � � �

� �

�

% 0 (2.5)12 3� �= 1.9734 V

Example 3.14.5

Solution : The dipole is shown in the Fig. 3.20

Given P (0.3, 0, 0.4) in cartesian co-ordinates.



Q1 Q2 Q30.5 0.5

d d

Fig. 3.19

� x = 0.3, y = 0, z = 0.4

� r = x y z2 2 2� � = 0.5

' = coszr

1� ��

��

= 36.8698º

# = tany

x1� = 0º

� Vp =Q

4d cos

r0 2%&'

��

��

=1.5 10

4 8.854 10

0.2 cos (36.8698º9

12

�

� �

��

�%

)

(0.5) 2

��

��

= 8.6282 V

��



x

y

z

(0, 0, – 0.1)

(0, 0, 0.1)r1

r2r'

P(r, , )' #

d = 0.2 m

Fig. 3.20


Example 4.2.5

Solution : a) From the continuity equation of current

I =

S

d� �J S = � �vol� ��J dv

As r = 3 m is constant, use surface integral.

dS = r d d2 sin � � �a r ... As J is in a r direction

J S�d =� �100 cos

sin�

� � �r 1

r d d2

2

S

d� �J S =� ��

�

�

� � � ��

� 0

100 cossin

r 1r d d

22

I =

��

�

��

� � �

� ��

�/ cos sin6 2100 r

r 1d d

2

2=

��

�

�

� �

� ��

�0

6 2/ sin100 r

r 1d d

2

2

=� �

� �100 r

2 r 1

2

2 ��

��

��cos 2

20

02�

��

�and r = 3 m

= � �50 910

2

2

0

22

��

� ��

��

�

��

�

�

��

cos cos�� = 70.6858 A

... use radian mode to calculate cos

b) Now r = 10 m and limits for � are 0 to �,

I =� ��

�

�

� �

� �� 0

2100 r

r 1d d

2

2

sin


4Conductors, Dielectrics

and Capacitance

(4 - 1)

70.6858 = � �50 r

r 1

2

2 ��

��

��cos 2

20

02�

��

�... Same I as before

70.6858 =� �

� ��

50 10

101

2

2

0

22

2��

��

��

��cos cos�

�

0.4545 = �cos 2 � + 1 i.e. cos 2 � = 0.5455

2 � = 56.9411� or 0.9938 rad i.e. � = 28.47� or 0.4969 rad

Example 4.2.6

Solution : From continuity equation,

I =

S

d� �J S

For y = 0 plane, the normal vector is a y and hence dS normal to a y is dx dz a y .

I =

z 0.002

0.002

x 0.1

0.1

x| dx dz

� � � �� 102| a ay y =

z 0.002

0.002

x 0.1

0.1

x| dx dz

� � � �� 102|

= � �10 22 z x | dx0.002

0.002

x 0

0.1

�� | = � �2 100 0 002 0 002� � � � �

�

��

�

��

. ( . )| x |

2

2

0

0.1

= 200 4 10(0.1)

23

2

� � � �� 4 mA

Example 4.2.7

Solution : J = � �12

3rA m2cos sin� � �a ar

i) I = J S�� d

S

... r = 20 cm is constant

dS = r d d2 sin � � � a r

J S�d =2

rcos r sin d d

3� � � �� 2 ... a a r� � = 0

I =

�

�

�

��

� ��

0

2

0

22 cos sin

rd d ... r = 0.2 m

� limits are from 0 to � 2 for the hemispherical shell

I = � �1

0 22

02

0

2

.sin�

�

!"#

��

�

�

d = � �1

0.2��

!"#

��

��

22

2 0

2

�� – cos

= � �10

20

��– cos cos = � �� 10

21 1

�– – = 31.41 A

ii) For spherical shell take � limits from 0 to �

Electromagnetic Theory and Transmission Lines 4 - 2 Conductors, Dielectrics and Capacitance


I =

�

�

�

� � ��

� ��

0

2

0

2 cos sin

rd d ... r = 0.1 m

= � �1

012

202

0.– cos�

�

!"#

��

��

��

�=

2 2 02

� �0.1

cos��

!"# �

��

��

– cos= 0 A

Example 4.2.8

Solution : The current is given by integral form of the continuity equation as,

I =

S

d� �J S

Nowd S = dr dz a � ... Normal to a � direction as J is in a � direction

J S�d = � � � �10 e dr dz100 r� �a a� �

= 10 e 100 r� dr dz ... � �a a� �� = 1

I =

S

100 re� �10 dr dz =

z 0

1

r 0.01

0.02100 re

� �

�� 10 dr dz

= � �10e

100z

100 r

0.01

0.02

01

�

�

�

��

�

��

= 10 ��

��

�

��

�

��

� �e100

e100

2 1[1]

= � �10 1.353 10 3.678 103 3� � � � � �� 2.326 10 2 A

Example 4.2.9

Solution : For z = 0, x = 1, from 3x + z = 3

For x = 0, z = 3, from 3x + z = 3

I = J S�� d

S

For x = 0, dS = dy dz ( )x�a

For x = 1, dS = dy dz ( )x a

For y = 0, dS = dx dz ( )y�a

For y = 2, dS = dx dz ( )y a

For z = 0, dS = dx dy ( )z�a

For z = 3, dS = dx dy ( )z a

I =

z 0

3

y 0

2

(x 0) z 0

3

y 0

2

3x dydz

� � � � �� 3x dydz (y 3) dxdz

(x 1) z 0

3

x 0

1

(y� � �

� �� 0)



x

y = 0

x = 0

z = 0

y = 2

y

x = 1

z = 3

z

Fig. 4.1

� � � � � ��

z 0

3

x 0

1

(y 2) y 0

2

x 0

1

(y 3)dxdz (z )dxdy (z 2)dxdy(z 0) y 0

2

x 0

1

(z

� � � � �

� �23)

= 0 3[y] [z] 3 [x] [z] [x] [z]02

03

01

03

01

03 � � 2 [x] [y] 5 [x] [y]

01

02

01

02

= 0 + 18 + 9 – 3 – 4 + 10 = 30 A

Example 4.2.10

Solution : The current is given by,

I =

S� �J dS

Assuming J given in a z direction, dS a zr dr d� �

J dS� =10

rr dr d 10 dr d

44� ��

I = � ��

�

�0

2

r 0

4 10 3

4 404 110 dr d 10 r

� �

� �� 0 3

02�

� �

= 10 4 10 24 3� � � � �� 80 A

Example 4.4.4

Solution : d = 0.8 mm, L = 2 cm, I = 20 A

| J| =

� �IS

I

4d

20

40.8 102 3 2

� ��

= 39.788 � 106 A/m2

Now | J| = $|E|

| E| =|J

0.686| 39.788 10

5.8 10

6

7$�

�

�� V/m

And V = EL = 0.686 2 10 2� � �� 0.0137 V

R =VI

0.013720

� � � �6.86 10 4 %

Example 4.4.5

Solution : L = 2000 ft = � �2000 30 10 2� � � m = 600 m

E =VL

1.2600

2 10 3� � � � V/m

a) J = $ E = 3.82 � � � �10 2 107 3 = 76.4 kA/m 2

b) I = JS = J d2��&

= 76.4 � �� 104

20 103 3 2�= 24 A

c) P = power dissipated = VI =VR

I R2

2� W = 1.2 24� = 28.802 W



Example 4.4.6

Solution : n = 1029 electrons / m 3 , E = 10 mV/m, $ = 5 107� S/m

i) J = $E = 5 10 10 107 –3� � � = 500 kA/m 2

ii) Area A =�4

d2 =�4

1 10 3 2� �( )– = 7.854 �10 7– m2

I = J A� = 500 10 7 854 103 7� � �. – = 0.3926 A

iii) ' e = ne = 10 1 6 1029 19� �(– . )– = –1.6 1010� C/m3 .

Note that e = Charge on one electron = – . –1 6 10 19� C

Example 4.6.4

Solution : ( e = 0.12, D = 1.6 n C m2

P = polarisation = ( )e 0 E

= ( )) )e 0

0 R

Dwhere ) (R e 1�

P =((

e

e

9

10.12 1.6 10

1.12D

��

= 0.1714 nC/ m 2

E =D

) )0 R

9

12

1.6 10

8.854 10 1.12�

�

� �

�

�= 161.3475 V/m

Example 4.6.5

Solution : For a dielectric medium,

D = ) )0 R E where )R = ( e 1 = 4.25 + 1 = 5.25

D = 8.854 10 5.25 0.15 1012 3� � � �� = 6.9725 10 C/ m15 2� �

and P = ( )e 0 E

P = 4.25 8.854 10 0.15 1012 3� � � �� = 5.644 10 C m15 2� �

Example 4.6.6

Solution : For the dielectric,

P = ( )�e E

Now )R = ( e 1 i.e. ( e = )R 1� = 2.8 – 1 = 1.8

And D = ) )� R E

E =D

) )0 R

7

12

3 10

8.854 10 2.8�

�

� �

�

�= 12.101 10 3� V/m

P = 1.8 8.854 10 12.101 1012 3� � � �� = 1.9285 10 C/ m7 2� �



Example 4.8.2

Solution :

i) M(4, –2, 1) is shown in the Fig. 4.2.

As point M is inside the conducter E D 0� � .

ii) N(–3, 1, 4) is shown in the Fig. 4.2. It is located

inside the dielectric.

DN = ' S i.e. DN = ' Sa y = 4 a y nC/m2

EN ='

) )S

0 r1

–9

–12

4 10

8.854 10 3�

�

� �= 150.59 V/m

Etan = Dtan = 0 � As per boundary conditions

E = EN = 150.59 a y V/m, D = DN = 4 a y nC/m 2

Example 4.8.3

Solution : E = 60 a x + 20 a y – 30 a z

D = � �)0–12= 8.854 10 60 20 – 30E a a ax y z�

D = 0.531 a 0.177 a – 0.265 ax y z nC/m 2

DN = D as Dtan = 0 as per boundary conditions

'S = � � � � � �D DN � � 0.531 0.177 0.2652 2 2 = 0.619 nC/m 2

Example 4.8.4

Solution : E = 0.2 a x – 0.3 a y – 0.2 a z V/m

Let the surface of the conductor is the interface between air and the conductor.

DN = )0 EN where EN = E

DN = 8.854 × 10– 12 [0.2 a x – 0.3 a y – 0.2 a z ]

= 1.7708 a x – 2.6562 a y – 1.7708 a z pC/m2

'S = DN = 1.7708 a x – 2.6562 a y – 1.7708 a z pC/m2

' S at that point = (1.7708) (2.6562) (1.7708)2 2 2 = 3.6506 pC/m2

Example 4.8.5

Solution : V = 10 (x2 + xy), P (2, – 1, 3)

i) V = 10 [22 + (2) (– 1)] = 10 [4 – 2] = 20 V

ii) E = – �V = –**

**

**

Vx

Vy

Vz

a a ax y z �

��

��



Conductor

MN

Dielectric

= 3)r1

y

z

Fig. 4.2

= – [(20x + 10 y) a x + 10x a y + 0 a z ]

EP = – (20 × 2 – 10) a x + 10 × 2 a y ] = – 30 a x – 20 a y V/m

iii) D = )0 E = 8.854 × 10– 12 [– 30 a x – 20 a y ]

= – 0.2656 a x – 0.1771 a y nC/m2

Example 4.9.6

Solution : The arrangment is shown in the Fig. 4.3.

For the boundary between two dielectrics,

DN1 = DN2

andE

EN1

N2=

))

2

1

Now D1 = 0.25 +C/m2

cos �, =D

DN1

1

cos 40� =D

0.25N1

DN1 = 0.1915 +C/m2

And DN2 = DN1 = 0.1915 +C/m2

Now D1 = � � � �D DN1 tan 12 2

0.25 = � � � �0.1915 D2tan 1

2



�1

�2

D E1, 1

Air

= 1)r1

Glass

= 4)r2 D E2, 2

Dtan1

Normal

DN1

Dtan2

DN2

40º

Boundary

Fig. 4.3

Dtan 1 = 0.1607 +C/m2

ButD

Dtan1

tan2=

))

))

,

� r1

r2

Dtan2 =))

r2

r1tan1D� � �

41

0.1607 = 0.6428 +C/m2

D2 = � � � �D DN2 tan22 2 = 0.6707 +C/m2

and cos � =D

D0.19150.6707

N2

2� = 0.2855

� = 73.41º

Example 4.9.7


From the boundary conditions of two dielectrics, D DN1 N2�

andE

EN1

N2=

))

2

1, �1 = 60º.

Nowtan

tan

��

1

2=

))

1

2=

))

r

r

1

2

tantan

60º�2

=72

�2 = 26.329º



�1

�2DN2

DN1

E D2 2,

D E1 1,

)r2 = 2

)r1 = 7

Normal

Dtan1

Dtan2

Fig. 4.4

Example 4.9.8

Solution : E1 = 100 80 60a a ax y z

At boundary, E1 = E Etan 1 N1

Now EN1 is projection of E1 in the normal direction, given by the dot product.

EN1 = � �E a a1 N12 N 12�= � �100 80 60

27

37

67

a a a a a a ax y z x y z N12 � ��

!"#

�

��

�� …

= � �� 28.57 34.285 51.428 57.143a aN12 N12

EN1 = – 16.326 a x + 24.489 a y + 48.979 a z V/m

Etan 1 = E E1 N1� = 116.326 a x + 55.511 a y + 11.021 a z V/m

At the boundary, Etan 2 = Etan 1 andE

EN1

N2=

))

,

EN2 =))

) )

,

�

�E EN1 N1�

3= 1.5 EN1

E2 = E E Etan 2 N2 tan 1 � 1.5 EN1 = 91.837 a x + 92.245 a y + 84.489 a z V/m

Example 4.9.9

Solution : As shown, z axis is normal to the surface. So part of E1 which is in the

direction of a z is normal component of E1 .

EN1 = 5a z V m

And E1 = E EN1 tan 1

Etan 1 = E E a a1 N1 x y� � �2 3 V/m

At the boundary of perfect dielectrics,

Etan 1 = E a atan 2 x y� �2 3 V/m

Now Dtan 2 = ) ) )�2 r2E Etan 2 tan 2�

and DN1 = ) ) )1 0E EN1 N1� r1

But DN2 = D EN1 N1� ) )� r1

and D2 = D D E EN2 tan 2 N1 tan2 � ) ) ) )� �r1 r2

= � � � �� )� 5 2 3 2 5a a ax y z� = 8.854 � �� 10 10 15 1012 a a ax y z

D2 = 88.54 a x � 132.81 a y 88.54 a z pC/m3

As D , EN1 N1 are in same direction and D , E1 1 are in same direction,

DN1 = D1 cos -�1 i.e. E EN1 1� -cos �1

where -�1 is angle measured w.r.t. normal.



EN1 = 5 and � � � � � �E1 � � 2 3 52 2 2 = 6.1644

cos -�1 =E

E5

6.1644N1

1� i.e. -�, = 35.795�

This -�, is angle made by E1 with the normal while �, is shown with respect to horizontal.

�, = 90 � -�, = 90 – 35.795 = 54.205�

Similarly if -� is angle made by E2 with the normal then,

cos -� =E

E

D

D

D

DN2

2

N2

2

N1

2� � =

) )� r1 E

D

N1

2... D DN2 N1�

=

� � � �

)

)

� � �

� � �

2 5

10 15 1002 2 2

1020.6155

= 0.485

-�2 = 60.982� i.e. � = 90 2� -� = 29.017�

Example 4.9.10


Fig. 4.5.

The electric field intensities are E1 and E2

in media 1 and 2 respectively.

As per boundary conditions,

Etan1 = Etan2

While J 1 and J 2 are current densities in the

two media. Similar to flux densities D the

boundary condition for current densities Jstates that J N1 = J N2 .

E

Jtan1

N1=

E

Jtan2

N2… (1)

From Fig. 4.5, J N1 = J 1 cos 1� and J N2 = J 2 cos 2� … (2)

From the point form of Ohm's law, J = $E

Etan1 =J tan1

$1and Etan2 =

J tan2

$2

From Fig. 4.5 J tan1 = J sin1 1� and J tan2 = J sin2 2�

Etan1 =J sin1 1�

$1and Etan2 =

J sin2 2�$2

… (3)

Using equations (2) and (3) in equation (1),



�1

�2

Boundary

Medium 1

$1

Medium 2

$2

J1

J2

( )E2

( )E1

JN1

Jtan1

Jtan2

JN2

Fig. 4.5

J sin

J cos

1 1

1 1

�$

�1

�

�

!

"#

=

J sin

J cos

2 2

2 2

�$

�2

�

�

!

"#

i.e.tan �

$1

1=

tan �$

2

2

tan

tan

��

1

2=

$$

1

2… Proved

Example 4.9.11

Solution : The normal direction to the y = 0, plane is a y hence out of E2 , 12a y is the

normal component of E2 .

EN2 = 12a y V/m

But E2 = E Etan 2 N2

Etan 2 = 5a ax z V m

At the boundary of perfect dielectrics,

Etan 1 = Etan 2 = 5a ax z V m

andE

EN1

N2=

))

r2

r1i.e.

EaN1

y12�

14

EN1 = 3a y

E1 = Etan 1 + EN1

= 5 a 3 a a V mx y z .

Example 4.13.5

Solution. : )r = 6, A = 0.254 x 0.254 cm2 , d = 0.254 cm

C =) )0 r A

d=

� �8.854 10 6 0.254 10

0.254 10

12 2 4

2

� � � �

�

� �

�= 0.135 pF

Example 4.13.6

Solution : )r � 3, A = 0.92 m2 , d = 4.5 mm

C =) ) )Ad

A

d8.854 10 3 0.92

4.5 10

0 r12

3� �

� � �

�

�

�

= 5.43 nF

Example 4.13.7

Solution : A = 1 cm 1 10 m2 4 2� � � , d = 1 cm = 1 10 2� � m, )r �6, )0128.854 10� � �

For parallel plate capacitor,

C =) ) )A

d

A

d6 8.854 10 1 10

1 10

r 012 4

� ��

�

� �

� 2= 0.5312 pF



y = 0

E1

)r1 = 4

)r2 = 1

E a a a2 x y z= + +5 12

Fig. 4.6

Example 4.13.8

Solution : A = 0.8 m2 , d = 0.1 mm, )r = 1000, E = 10 V m6

C =)Ad

=) )0 r A

d=

8.854 10 1000 0.8

0.1 10

–12

–3

� � �

�= 70.832 F+

E =Vd

i.e. 106 =V

0.1 10–3�i.e. V = 100 V

Example 4.13.9

Solution : A = 100 cm2, d = 2 mm, C = 2 10 4� � +F, V = 20 kV

a) C =QV

i.e. 2 10 104 6� �� =Q

20 10 3�i.e. Q = 4 +C

The flux is same as the charge, . = Q = 4 +C

b) E =Vd

=20 10

2 10

3

3

�

� �= 10 106� V/m =

10 10 kV100 cm

3�= 100 kV/cm

c) C =) )0 r A

di.e. 2 10 104 6� �� =

8 854 10 100 10

2 10

12 4

3

. � � �

�

� �

�

)r

)r = 4.5177

d) D = ' S =QA

=4 10

100 10

6

4

�

�

�

�= 4 10 4� � C/m2

Example 4.14.2

Solution : The capacitance of a co-axial cable is,

C =2 �) L

ba

ln��

!"#

where a = Inner radius, b = Outer radius

a =0.0295

2= 0.01475 inches and b =

0.1162

= 0.058 inches

As ratioba

is to be used, no need to convert to metres.

C =� �2 8.854 10 2.26 20 10

0.0580.01475

12 2��

��

!"#

� �

ln

= 18.365 pF

Example 4.14.3

Solution : The D field between the plates in cylindrical co-ordinates is of the form

D = D� �a where D� depends only on r.

Let plate at � = 0º is V = 0 and plate at � = �º is V = Vo.

Vo = – dE L� �� –D

rd

0

��

�

�

)�a a

�

�

!

"##�

��



=–D r

d–D r

0

�

�

��

)�

�

)�� Where ) = ) )0 r

D� = –)�

Vr

o

The charge density on the plate � = � is

'S = Dn = – D� =)�

Vr

o

The total charge on the plate is,

Q = ' Sds� =)

�V

rdr dzo

r r1

r2

z 0

L

�� =

)�

V L r

ro 2

1ln

C =QVo

=)�L r

r2

1ln

Example 4.15.5

Solution : The sphere is shown in the Fig. 4.7

a = Radius of sphere =d2

= 1 cm

r1 = a + Thickness = 1 + 3 = 4 cm

This forms two capacitors in series.

C1 = Capacitor due to spherical arrangement of

concentric spheres

=4

1a

1b

�)

��

!"#

Here a = 1 cm and b = r1 = 4 cm

C1 =4 2.26

1

1 10

1

4 10

3.35270

2 2

� )� �

�

�

�

!

"# �

�

�

�

!

"#

�

��

�

��

�

� �

pF

And C2 = Due to isolated sphere of r1 = 4 cm = 4 �) r1

But between this sphere of radius r1 and sphere at /, the dielectric is free space )0 .

C2 = 4 0�) �r 4 8.854 10 4 10 4.4505 pF112 2� � � � � ��

Ceq =1

1C

1C1 2

... C1 and C2 in series

=C C

C C1 2

1 2=

� �3.3527 10 4.4505 10

10 3.3527 4.4505

12 12

12

� � �

� �

�

= 1.9121 pF



a

r1

3 cm

Fig. 4.7

Example 4.15.6


As )r1r

0 , the standard formula for spherical capacitor

can not be used.

In spherical conductor E at a radial distance r is given

by,

E =Q

rV m

24 �)a r

V = � ��

� E Ld

= � ��

�

�r 10 cm

r 5 cm

2

Q

rdr

4 �)a ar r ... Note ) � ) )� r

= ��

��

��

�

��

Q 1

0.1r

r

drQ

r 0.1

r 0.05

2 r4 4�) �)� � 0.1

0.0510r� dr

= � �1 2� � � ��

��

10

4

10

4

Qr

Q 0.050.10.1

0.05

�) �)� �ln ln ... Q = 1 +C

= 62.298 kV

And C =QV

1 10

62.298 10

6

3�

�

�

�= 16.051 pF

Example 4.15.7

Solution : For spherical capacitor,

C =4�)

1a

1b

��

��

and a = 1 cm, b = 2 cm and )r = 2

C =4 2 8.854 10

1

1 10

1

2 10

12

2 2

��

��

�

�

��

�

��

�

� �

= 4.4505 pF

Example 4.15.8

Solution : i) The capacitance of a single isolated sphere is,

C = 4 �) a ... ) )� 0 , a =1.52

0.75 m�

= 4 8.854 10 0.7512�� 83.447 pF



++

+

+

++

+

–

–

–

––

–

–

–

) 0r1–r

5 cm

10 cm

Fig. 4.8

ii) For co-axial cable,

C =2 L

ba

�)

ln ��

��

where, L = 1.5 m, )r = 2.26, a = 0.6 mm, b = 3.5 mm

C =2 8.854 10 2.26 1.5

3.50.6

12��

��

��

��

ln

106.935 pF

iii) Consider the cylindrical conductor suspended above the conducting plane as shown in

the Fig. 4.9

The capacitance of this arrangement is given by,

CL

=2

coshhb

1

�)

� ��

!"#

) = )0 , h = 15 m, b = 1.5 10 m3� �

CL

=2 8.854 10

cosh15

1.5 10

12

13

��

�

�

�

!

"#

��

��

5.6173 pF m

Example 4.16.6

Solution : In this case, the dielectric interface is normal to the plates. Hence two

capacitors formed are in parallel.

Now d = 2 mm same for both the dielectrics

A 1 = 1.2 2 2.4 m2� � and A 0.8 2 1.6 m22� � �

C1 =) ) )

)0 R1 112

R13

8R1

A

d

8.854 10 2.4

2 101.062 10 F�

� �

��

�

��

C2 =) )0 R2 2

12

38A

d8.854 10 1

2 101.7708 10 F�

� � �

��

�

��25 6. .

Ceq = C C 1.7708 10 1.062 101 28 8

R1 � � �� )

But Ceq = 60 nF

60 10 9� � = 1.7708 1 1.062 108 8R1� �� 0 ) i.e. 1.062 10 8

R1� � ) = 4.2292� �10 8

)R1 = 3.9823



h = 15 m

b = 1.5 mm

Plane

Fig. 4.9

Example 4.16.7

Solution : a) The arrangement is three capacitors in series.

C3 =3 8.854 10 20 10

0.4 10

12 4

3

� � � �

�

� �

�

= 132.81 pF

C2 =2 8.854 10 20 10

0.3 10

12 4

3

� � � �

�

� �

�= 118.053 pF

C1 =1 8.854 10 20 10

0.2 10

12 4

3

� � � �

�

� �

�= 88.54 pF

1

Ceq=

1C

1C

1C1 2 3

hence Ceq = 36.6372 pF

b) Let the supply voltage be V.

ET = Total energy =12

C Veq2

E1 =12

C V1 12 , E2 =

12

C V2 22 , E3 =

12

C V3 32

For capacitors in series, Q = C1 V1 = C2 V2 = C3 V3 = Ceq V

ET =12

QC

2

eq, E1 =

12

QC

2

1, E2 =

12

QC

2

2, E3 =

12

QC

2

3

% Energy stored in region 1 =C

Ceq

1�100 = 41.3792 %


Ceq

2�100 = 31.0345 %


Ceq

3�100 = 27.5863 %

Example 4.16.8

Solution : Refer Example 4.16.3 for the produce hence

Ceq = 52.425 pF , V = 200 V, Q = C V1 1 = C V2 2 = C Veq

V1 = 105.2632 V, V2 = 94.7368 V

E1 =V

d1

1= 26.3158 kV m, E2 =

V

d2

2= 15.7894 kV m

Example 4.16.9


d2 = 0.002 m

d = d d1 2 = 0.01 m

d1 = 0.01 – 0.002 = 0.008 m



C = A/d3 R3 0 3) )

C = A/d2 R2 0 2) )

C = A/d1 R1 0 1) )

Fig. 4.10

)r2 = 6 r1) = 6

A = 1 m2

C1 =)1

1

A

d

and C2 =)2

2

A

d

and two are in series.

C1 =) )0 r1

1

12A

d8.854 10 1 1

0.008�

� � ��

�1.1067 10� 9 F

C2 =) )0 r2

2

12A

d8.854 10 6 1

0.002�

� � ��

�26.562 10� 9 F

Ceq =C C

C C1.1067 10 26.562 10

1.1067 10

1 2

1 2

9 9

�

� � �

�

� �

� 9 926.562 10 � �

= 1.0624 10 9� � F = 1.0624 nF

Example 4.17.6

Solution : d = 5 mm, S = 80 cm2, )r = 10

i) C =) S

d=

8.854 10 10 80 10

5 10

12 4

3

� � � �

�

� �

�= 141.664 pF

ii) C =QV

i.e. Q = CV = 141.664 � 10– 12 � 50 = 7.0832 nC

E =Vd

=50

5 10 3� �= 10 kV/m, D = ) )0 r E = 0.8854 µC/m2

WE =12

CV2 =12

� 141.664 � 10– 12 � (50)2 = 0.17708 µJ

iii) Though source is disconnected and the dielectric is removed, Q on the surface remainssame

Q = 7.0832 nC, D = ' S =QS

=7.0832 10

80 10

9

4

�

�

�

�= 0.8854 µC/m2

E =D

0)=

0.8854 10

8.854 10

6

12

�

�

�

�= 100 kV/m … )r = 1 as dielectric removed

WE =12

CV2 =12

CQC

��

��

2

=12

QC

2

But C =)0 S

d=

8.854 10 80 10

5 10

12 4

3

� � �

�

� �

�= 14.1664 pF



d2

d1

Wood

Air

)r1 = 1

)r2 = 6

Fig. 4.11

WE =12

(7.0832 10 )

14.1664 10

9 2

12

�

�

�

�= 1.7708 µJ

iv) V =QC

=7.0832 10

14.1664 10

9

12

�

�

�

�= 500 V

Example 4.17.7

Solution : A = 36 10 4�� m2, V1 = 100 V, d1 = 1 mm = 1 10 3� � m, )r = 1

C1 =) )0

1

r A

d= 100.1363 pF

E1 = Energy stored =12

C V1 12 = 0.5 +J

and Q = C1 V1 = 1 10 8� � C

Now battery is disconnected, but charge Q remains same.

d2 = 2 mm = 2 10 3� � m, New C = C2, New p.d. = V2

Q = C2 V2 and C2 =) )0

2

r A

d= 50.068 pF

1 × 10 8� = 50.068 10 12� � V2

V2 = 199.7283 V 3 200 V

E2 =12

C V2 22 = 0.998 +J 3 1 +J

i) Change in p.d. = 100 V, V2 > V1ii) Change in energy stored = 0.5 +J, E2 > E1

Example 4.17.8

Solution : i) The capacitance per unit length is given by,

CL

=2�) )o r

lnba

��

��

and b = 2a (given)

CL

=2 8 854 10 25 1012 6�� .

ln[2]

= 2 × 10– 15 F/m

So for L = 1 m, C= 2 × 10– 15 F

ii) E =12

CV2 =12

× 2 × 10– 15 × (2 × 105)2 = 40.129 µJ

Example 4.17.9

Solution : a) C =) A

d8.854 10 12 120 10

5 10

12 4

3�

� � � �

�

� �

�= 255 pF



b) C =QV

Q = CV = 255 10 4012� �� = 10.2 nC

E =Vd

40

5 10 3�

� �= 8 kV/m

D = ) E = ) )� R E = 0.85 +C/m 2

WE = � �12

CV12

255 10 402 12 2� � � �� = 0.204 +J

c) Though source and dielectric is removed, Q on the surface remains same.

Q = 10.2 nC

D = ' S

9

4

QA

10.2 10

120 10� �

�

�

�

�= 0.85 +C/m 2

Now E =D)�

= 96 kV/m ... Now ) � )�

and as V is now not same across the plates, calculate WE as,

WE =12

CV12

CQC

22

� ��

��

... As VQC

�

WE =12

QC

2

and C =)0 A

d= 21.249 pF ... ) � )�

WE =� �1

2

10.2 10

21.249 10

9 2

12

�

�

�

�= 2.448 +J

d) V =QC

10.2 10

21.249 10

9

12�

�

�

�

�= 480 V

Example 4.17.10

Solution : C1 =)1

1

A

dand C2 =

)2

2

A

d

And C1 and C2 are in series.

C1 =) )0 r1

1

A

d=

8 854 10 3 100 10

3 10

12 4

3

. � � � �

�

� �

�= 8.854 10 F11� �

While C2 =) )0 r2

2

A

d=

8 854 10 2 100 10

2 10

12 4

3

. � � � �

�

� �

�

= 8 854 10 11. � � F



Ceq =C C

C C1 2

1 2= 4 427 10 11. � � F

Now Ceq =QV

Q = Ceq V = 4 427 10 10011. � ��

= 4 427 10 9. � � C

The charge Q remains same for C1 and C2

C1 =QV1

hence V1 = 50 V

While C2 =QV2

hence V2 = 50 V

The energy stored in each dielectric is,

WE1 = WE2 =12

C V1 12 =

12

C V2 22 = 0.1106 +J

The potential gradients are,

E1 =V

d1

1=

50

3 10 3� �= 16.667 kV/m

and E2 =V

d2

2=

50

2 10 3� �= 25 kV/m

Example 4.17.11

Solution : A = 1 m2 , d = 1 mm, )r = 25, V = 1000 V

C =) ) )Ad

A

d0 r� =

8.854 10 25 1

1 10

12

3

� � �

�

�

�= 221.35 nF

Q = CV = 221.35 10 1000 2.2135 10 C9 4� � � ��

For the plate separation 'x', the capacitor is C =)Ax

.

For the fixed voltage V across the plates,

VQx

**

= V(CV)

xV

Cx

2**

�**

While the energy stored in a capacitor =12

CV2

**W

xE =

**

��

��

�**x

CV VCx

2 212

12

The force between the plates is given by,

F = �**

**

� �**

**

W

xV

Qx

VCx

VCx

E 2 212



d2

d1

)2

)1

A

Fig. 4.12

=12

VCx

2 **

but C =)Ax

hence**Cx

=�)A

x2

F = �12

VA

x

22

)= �

� � � �

� �

�

�

(1000) 8.854 10 25 1

2 (1 10 )

2 12

3 2= – 110.675 N ... x = d = 1 mm

Example 4.17.12

Solution : A = 30 � 30 = 900 cm2 , V = 1000 V, d = 5 mm, )r = 1

C =) )0

12 4

3

8 854 10 1 900 10

5 10

r A

d�

� � � �

�

. – –

–= 159.372 pF

WE =12

CV2 =12

159 372 10 100012 2� � �. ( )– = 79.686 +J

| |E =Vd

=1000

5 10 3� –= 200 10 3� V m

Energy density =12 0

2) )r| |E =12

8 854 10 200 1012 3 2� � � �. ( )–

= 0.17708 J m3

��




Example 5.2.5

Solution : Find � 2 V

� 2 V =1 1 12

22 2r r

rV

r r

V

r

��

��

��

��

�

��

� �

��

�

sinsin

sin 2

2

2�

�

� �

V

=1

502 1 50

2 3 2 2r r r r r

��

��

��

�r 2 ��

��

�

� sin

( )

sinsin cos ��

��

�

� 0

=1 100 1 502 2 2r r r r r

��

�

�

��

� ��

��

�

��

�

sin

sin

sin cos

=1

1001 1 1 50

22

2 2 2 2r r r r( sin )

sinsin� �

�

��

�

��

��

�

��

��

=�

� 100 1

25 2 24 4

sin

sincos

�

��

r r=

100 150 1 2

4 42sin

sin[ sin ]

�

��

r r� �

=100 50 100

4 4 4

sin

sin

sin�

�

�

r r r� � =

500

4r sin ��

Hence given potential field does not satisfy Laplace’s equation.

Example 5.2.6

Solution : In cartesian system let the vector A is,

A = A A Ax y za a ax y z� �

� 2A = � � � � �2x

2y

2zA A Aa a ax y z

� L.H.S. = � �� 2A = � �� 2x

2y

2zA A Aa a ax y z

=�

�

�

��

��

��

��2

x2

y2

zA

x

A

y

A

z

=��

�

�

��

�

�

��

�

�xA

x y

A

y zA

z

2x

2

2y

2

2z

2

�

��

�

�

�

�

��

�

�

�

��

�


5 Poisson's and Laplace's Equations

(5 - 1)

R.H.S. = � �� 2 A = � � ��

��

�

2 ��

�

��

Ax

A

yA

zx y z

=�

�

��

�

�

�

��

�

�

�

2

2x

2

2

y 2

2z

x

Ax y

A

y z

A

z

��

��

�

��

�

� �

��

=��

�

�

��

�

�

��

�

�xA

x y

A

y zA

z

2x

2

2y

2

2z

2

�

��

�

�

�

�

��

�

�

�

��

�

= L.H.S. ... Proved

Example 5.2.7

Solution : V = 2 xy2z3

� VP = 2 × 1 × (3)2 × (– 1)3 = – 18 V

The Laplace's equation is � 2 V = 0

� 2 V =�

�

�

�

�

�

2

2

2

2

2

2

V

x

V

y

V

z� � =

��

��

��x

[2y z ]y

[4xyz ]z

[6xy z ]2 3 3 2 2� �

= 0 + 12 xz3 + 12 xy2z

As � 2 V � 0, the given V does not satisfy Laplace's equation.

Example 5.4.9

Solution : The spherical shells are shown in the Fig. 5.1.

The E is in radial direction and hence V is also the function

of r alone and independent of � and �.

�� 2 V =1

r rr

Vr

02

2��

��

��

�� ... Laplace's equation

��

��

��

��r

rVr

2 = 0

Integrating, rVr

2 ��

= � �0 C = C1 1 ... (1)

��Vr

=C

r

12

Integrating, V = � �C

rdr + C =

C

r+ C1

2 21

2 ... (2)

At r = 0.1 m, V = 0 and r = 0.2 m, V = 100 V

� 0 = �C

0.1+ C1

2 and 100 = �C

0.2+ C1

2

Solving, C1 = 20, C2 = 200

� V = �20r

+ 200 V

Electromagnetic Theory and Transmission Lines 5 - 2 Poisson's and Laplace's Equations


V = 100 V

r = 0.2m

r = 0.1m�0

Fig. 5.1

Hence E = ��

V =Vr

a r = ��

� ��

�

� � ��

��

�

��

� !

"#$r

20r

200 201

r 2a ar r

� E = �20

ra

2 r V/m

� D = ��

0020

E a177.08

rar 2 r�

��

r 2pC/m 2

Note that as outer shell is at higher potential, E is directed from outer to inner shell andhence in �a r direction.

Example 5.4.10

Solution : The potential is changing with respect to y only hence,

� 2 V =�

�

2

2

V

y= 0 ... Laplace's equation

Integrating twice, V = C y C1 2�

� E = ��

��

��

�

��

V

Vx

Vy

Vz

a a ax y z = � C1 a y

� D = � �0 1 0CE a y�� C/m2

But given D = 253 a Ny C/m2 i.e 253 10 9 � = � C1 0�

� C1 = �

� �

�

��253 10

8.854 1028.574 10

9

23

1

At y = 0.01 m, V = 0 i.e. 0 = � �28.574 10 0.01 C32

� C2 = 285.746

� V = � �28.574 10 y 285.7463

So voltage at y = 0 m is, V = 285.746 Vand voltage at y = 0.02 m is, V = – 285.746 V

Example 5.4.11

Solution : Refer similar example 5.4.6 in which % � &' 6 and Vo = 100 V.

� V =� �

V 100

/ 6

600Vo

%�

&�

&��

E =–Vr

–100

r6

– 600r

o

% & &� � �a a a��

��

�

Example 5.4.12

Solution : V is a function of � only and not the function r and z.



� � 2 V =1

r

V2

2

2

�

�� 0 ... Laplace's equation

Integrating,��V

= � � �0 d C = C1 1

Integrating, V = � � �C d + C = C + C1 2 1 2

At ��()*, V = 50 V while at �� )3 *, V = 20 V

� 50 = 10 C + C1 2 and 20 = 30 C + C1 2

Subtracting, 30 = – 20 C1

� C1 = – 1.5 and C2 = 65

� V = – 1.5 � + 65 V ... Use � in degrees.

Now at P (2, 1, 3), x = 2, y = 1, z = 3

� tan � =y

x�

12

� � = 26.56*

� VP = – 1.5 *�26.56 65 = 25.152 V

Example 5.4.13

Solution : V is a function of � only and not the function of r and � .

� � 2 V =1

r sin

V2 � �

��

��

��

��

�� sin 0 ... Laplace's equation

��

��

��

��

��

sinV

= 0

Integrating,

sin ��

��V

= � 0 d + C = C1 1� i.e.��V�

=C

C cosec11sin �

��

Integrating, V = � C cosec d + C1 2� � = C C1 2ln tan�2

��

��

At � �+)*, V = 50 V and at � � )5 *, V = 20 V

� 50 = C + C1 2ln tan302*�

��

and 20 = C + C1 2ln tan502*�

��

i.e. 50 = – 1.3169 C C1 2� and 20 = – 0.7629 C C1 2�

Subtracting, 30 = – 0.5539 C1

� C1 = – 54.152, C2 = – 21.3125



� V = � ��

��54.152 tan

221.3125 Vln

�... Use � in degrees.

For P (2, 1, 3), x = 2, y = 1, z = 3

� � = cos cos cos� � ��

��

�

��

�

�

��

1 1 1 3

14

zr

z

x y z2 2 2= 36.6692*

� VP = �*�

��54.152 tan

36.69922

21.3125ln = 38.4489 V

Example 5.4.14

Solution : At origin, V(0, 0) = f(0) = 0

E = ��

��

�

��V

Vx

Vy

Vz

��

��

��

a a ax y z = � ��

�

� ��

��

15x

f xx

4y 02 � � ��

a a ax y z

� Ex = – 15xf(x)

x2 ��

��

��

At origin, Ex = 0 hence��f(x)

x0� at origin

For a charge free region, � �2 V 0

��

�

�

�

�

�

2

2

2

2

2

2

V

x

V

y

V

z� � = 0

��Vx

= 15xf xx

2 ��

,��Vy

= – 4y,��Vz

= 0

��

�

2 V

x2= 30x

f x

x2��

�

2,

��

,Vy

= – 4,�

�

2

2

V

z= 0

� 30 xf x

x4 0

2

2� � ��

�= 0 i.e.

�

�

, f(x)

x2= 4 – 30x

Integrate,��f(x)

x= � � � � � �(4 30 x)dx A 4x

30x2

A1

2

1

But at origin,��f(x)

x= 0 i.e. 0 = A 1

Integrate, f(x) = � ��

��

�

�� 4x

30xdx A

2

22=

4x2

30xA

2 3

2� �6

But f(x) = 0 at origin hence 0 = A 2

� f(x) =4x

230x

6

2 3� � �2x 5x2 3



Example 5.4.15

Solution : As -v � 0, use Poisson's equation

� 2 V =� �

� � �� -

� �v 0

0

x2 107

... � �� 0 as free space

Now V is a function of x alone, hence ��

�2 V =

V

x

2

2.

��

�

2

2

V

x= 2 107 1 2 x /

Integrating,��Vx

=2 10

32

7 3 2 � � �

xC 13.33 10 x C1

6 1.51

/

Integrating, V = � �13.33 10 x + C dx C6 1.51 2 �� =

� �13.33 10 x

2.5+ C x C

6 2.5

1 2

�

� V = 5.33 10 x C x C6 2.51 2 � �

At x = 0, V = 0 hence 0 = 0 + 0 + C , C 02 2 �

At x = 2.5 mm, V = 2 V hence

2 = � � � �5.33 10 2.5 10 C 2.5 106 3 2.5

13 � � �

� C1 = 133.75

� V = 5.33 10 x + 133.75 x V6 2.5

At x = 1 mm = 1 10 m3 � ,

V = � � � �5.33 10 1 10 + 133.75 1 106 3 2.5 3 � � = 0.30229 V

Example 5.4.16

Solution : The two cones are shown in the Fig. 5.2.

The potential is constant with r and � and is the function of � only.

So Laplace's equation reduces to,

1

r sin

dd

sindVd2 � �

��

��

�

= 0

Integrating, sin ��

dVd

= 0 A�� = A

IntegratingdVd�� =

Asin �� d� + B = A cosec�� B



� V = A ln tan�2

��

��

�

��

+ B

For �1 =&

10, V1 = 0 V

� 0 = A ln tan/& 102

��

��

�

��

+ B i.e. 0 = – 1.8427 A + B … (1)

For �2 =&6

, V2 = 50 V

� 50 = A ln tan/& 62

��

��

�

��

+ B i.e. 50 = – 1.3169 A + B … (2)

Solving equations (1) and (2), A = 95.09319, B = 175.2282

V = 95.09319 ln tan�2

��

��

�

��

+ 175.2282

E = – � V = –1r

dVd� �a

…Other terms are zero

= –1r

dd

95.09319 tan2

175.2282�

�ln �

��

��

�

��

� !

"#$

a �

= –1 1 1

2r95.09319

tan2

sec2

2 ��

��

��

��

�

..

!..

"

#.

�� .

$..

a �

= –95.09319

r

cos2

2sin2

1

cos2

2

��

��

��

��

��

��

�

� �a �

= –95.09319

r1

2sin2

cos2

��

��

��

��

� � �a

= –95.09319

r sina

� � V/m

��



V = 0 V1

V = 50 V2

�1

&� = —

101

&� = —

62

�2

Fig. 5.2


Example 6.3.4

Solution : The points are shown in the

Fig. 6.1

The direction of dL is from A to B. So let

us obtain unit vector in the direction from

A to B.

aAB =R

RAB

AB| |

=( ) ( ) ( )

( ) ( ) ( )

� � � � � � �

� � � � �

2 1 1 2 3 4

3 3 12 2 2

a a ax y z

=� � �3 3

19

a a ax y z

� dL =� �

dL aa a a

ABx y z

�� 10 3 3

19

4

Now, a R12 =R

R

a a a12

12

x y z�

� � � � � �

� � � �

( ) ( ) ( )

( ) ( ) ( )

3 1 1 2 2 4

2 1 62 2 2=

2 6

41

a a ax y z� �... From A to C

� dL a R12� =

a a a

3 3 1

3 1 6

x y z

� � �

� �

� � �17 20 9a a ax y z

... (without considering1

19

1

41and )

� I dL a R12� =� �6 10 17 20 9

19 41

4� � �

�

� a a ax y z= � �2.1497 10 5� � �� 17 20 9a a ax y z


6 Magnetostatics

O

x

y

aR12

R12

dL

Point 1A (1,2,4)

Point 2C (3,1,–2)

B (–2,–1,3)

Fig. 6.1

(6 - 1)

� dH =� �I d

(R )

2.1497 10

122

5L a a a a

R12 x y z��

� � �

�

�

4

17 20 9

4 41 2 ( )

= � �4.172 10 8� � �� 17 20 9a a ax y z

= 0.7093 a 0.8344 a 0.3755 ax y z� � A/m

Example 6.3.5

Solution : P(0, 1, 0) is a point at which H is to be obtained.

Case 1 : �� z hence general point on the element is (2, – 4, z).

R12 = � �� 0 – 2 1 – – 4 0 z 2 za a a a a ax y z x y z� � � � � �– 5

| |R12 = 4 25 29� � � �z z2 2 , aR

RR1212

12�

| |

dHP =I

4 R122

dL a R12�

= � �

0.4dz –2 5 – z

4 z + 29 z2 2

a a a az x y z� �

� 29

a a ax y z

0 0 0.4

2 5 z� �

= – 0.8 – 2a ay x

� dHP =

��

�

�

�

��

�

�

��

0.8 2

4 z 29

dz2 3/ 2

a ay x

i.e. HP =

��

��

�

�0.8 2

4 z 29

dz2 3/ 2

z –

a ay x

Putting z = 29 tan �,

�dz

z 29

=1

29d =

2

292 3/ 2z – � � �

�

� �

�

�� cos

º

º

� �� 90

90

Changing limits of integration from z = 29 tan �.

� z = – �, � = –90º and z = +�, � � 90º

� HP = � �2

4 29– 0.8 – 2

�� a a 0.0591 a – 0.0236 ay x x y A/m

Case 2 : � � �3 3z

Only change is the limits of integration.

�

�dz

z 292 3/ 2z –3

3

�� =

1

29cos d

–5.906º

5.906º

��

� � �0.2058

29

� HP = � �0.2058

4 29– 0.8 – 2

��a a – 0.00608 a – 0.00243 ay x x y A/m

Electromagnetic Theory and Transmission Lines 6 - 2 Magnetostatics


Case 3 : 0 < z < �

�

�dz

z 292 3/ 2z 0 ��

�

� =1

29

1

290

90

cos

º

º

� �d ��

� HP = �1

4 29– 0.8 – 2

��a a – 0.0295 a – 0.0118 ay x x y A/m

Example 6.3.6

Solution : The small wire is shown in the Fig. 6.2.

a RQP =R

RQP

QP| |

RQP = (0 1) (2 0) (2 0)x y z� � � � �a a a

= � � �a a ax y z2 2

|RQP | = 1 4 4� � = 3

� a RQP =� � �a a ax y z2 2

3

I dL = 2a x

� dH =I

4 R

RQP

2

dL a�

I RQPdL a� =

a a a

2 0 013

23

23

x y z

�

=43

a az y43

�

� dH =

43

4 3 2

a az y43

�

� ( )= � 0. 0117 a + 0 . 0117 ay z A/m

Example 6.3.7

Solution : The square is placed in the xy plane as shown in the Fig. 6.3. (see figure on

next page)

Consider differential element dx along AB of the square.

� dL = dx a x

The R12 joining differential element to point P is,

R12 = – x a x – y a y i.e. R12 = x y2 2� i.e. a R12 =� �

�

x y

x y2 2

a ax y

� dL a R12� =

a a a

dx 0 0

x y 0

y dxx y z

� �

� � a z



O

Q(1, 0, 0)

RQP

z

y

x

P(0, 2, 2)

ax

2A

Fig. 6.2

According to Biot-Savart law,

dH = �I d

R

I y dx

x y x y122

2 2 2 22

L a aR12 z��

�

� ��

��

44

... Considering R12

=

�10 ( 2.5) dx

x 2.52 2 3/ 2

� �

�

a z

4 ... y = 2.5 for segment AB

� H =

�x 2.5

2.5

2 2 3/ 2x 2.5

025 dx

4 x 2.5

2

�

�

��

�

��

�a z

2

�5 dx

x 2.52 2 3/ 2

a z

4 �

Put x = 2.5 tan �, dx = 2.5 sec2 � d�

Limits, x = 2.5, � � �� and x = 0, �� 0

� H =

� ��

�

��

�

�

�25 2

4145

0

2 3 2� �

��

2.5 sec d

2.5

0.63662

3

a z

tan/

��

��

�

�45

01

secd a z

= – 0.6366

�

� ��

�

�45

0

cos d a z = – 0.6366 � �sin �450

�� a z = – 0.6366 � �0 45� �sin a z

= 0.4501 a z A/m

This H is due to the segment AB of the square. All sides will produce same H at point P.

� Htotal = 4 4H � � 0.4501 a z = 1.8 a z A/m

Example 6.3.8

Solution :R12 = � � �3 4 5a a ax y z



P

Point 2

A

BC

D

R12

2.5

2.5

–2.5

–2.5

dxPoint 1

y

x

z

dx

P

–yay

–xaxR12

Fig. 6.3

� a R12 =R

|R |12

12=

� � �

� �

3 4 5

3 4 52 2 2

a a ax y z

= � � �0.4242 0.5656 0.7071a a ax y z

� dH2 =I

R

1

12

2

dL a1 R12�

4

I 1dL1 = I (given)1dL = 3 2 3 ( )a a ax y z� � Am

I 1dL a1 R12� =

a a ax y z

3 6 9

0.4242 0.5656 0.7071

� � �

= � � � � � �13.328 5.33 6.664a a a a a ax y z z y x12 8 16

= 2.672 5.336 2.67a a ax y z� �

� dH2 =2.672 5.336 2.67a a ax y z� �

�4 50 2 ( )= 4.252 a 8.4925 a 4.252 a nA mx y z� �

Example 6.4.4

Solution : The magnetic field strength due to

straight conductor carrying current I, at a point at

a distance of r from it is given by,

H =I

2 ra

r1 = (0.75) (0.75)2 2� = 1.0606

r2 = (0.75) (0.25)2 2� = 0.7905

r3 = (0.75) (0.25)2 2� = 0.7905

Key Point All the currents are coming out of paper hence according to right hand thumb

rule, produce H in the same direction at P.

� HP = H H H1 2 3� � =I I I1 2 3

2 2 21 2 3 r r ra a a� �

=50

210

240

2 ��

��

��

��

��1.0606 0.7905 0.7905a

= 17.5697 a A/m



0.5 m 0.25m

0.25m

0.75

P

!3

r1 r2r3

!1 !2Fig. 6.5

R12

IdL

P(3,4,5)

O(0,0,0)

Fig. 6.4

Example 6.5.5

Solution : i) P (2, 2, 0) : Consider the four sides separately as shown in the Fig. 6.6.

� HP = � � � �Ir

Ir4 41

2 12

2 1" "

" "sin – sin sin – sin�

� � � �� Ir

Ir4 43

2 14

1 2" "

" "sin – sin sin – sin

Hp =Ir

Ir

Ir4 4 41

2 12

2 13

2" "

" "

"[sin – sin ] [sin – sin ] [sin –� � sin ]"1

�Ir4 4

1 2" "[sin – sin ]

= � �10

4

sin 45 – sin –45

2

sin 45 – sin –71.56

2�

#$% � � � �

� �&'(

sin 18.43 – sin –18.43

6

sin 71.56 – sin –45

2

= ) *104

0.7071 0.8278 0.1053 0.8278

� � � = 1.964 a A mz



r = 21

2

P(2,2,0)I = 10 A

2

"2"1

2

"

"2

1

= 45º

= –45º

(a) (b)

6= –tan –— = –71.56º

2"1

–1

P(2,2,0)

2 6

"2"1

r = 22

I = 10 A

"2 = 45º

2= –tan –— = –18.43º

6"1

–1

P(2,2,0)

r = 63 I = 10 A

2

2

"1"2 6

2= tan –— = 18.43º

6"2

–1

(c) (d)

6= tan –— = 71.56º

2"2

–1

P(2,2,0)

2 6

"2"1r = 24

I = 10 A

2= –tan –— = – 45º

2"1

–1

Fig. 6.6

Use the above procedure for the remaining points and verify the answers :

ii) 1.78 a A mz iii) –0.1178 a A mz iv) –0.3457 a – 0.3165 a 0.1798 a A mx y z�

Example 6.5.6

Solution : Consider one side of a square as

shown in the Fig. 6.7, in xy plane. Consider

segment BC, which is finite length of the

conductor. As B is above P, "+ is negative and

", is positive.

"+ = tan2.52.5

451� � � , but

"+ = � �45 and ", = � �45

� H =I

r4 " ", +sin sin�

= �� 104

45 45�

� � � �2.5

sin sin ( )

= 0.4501 A/m

Important note : Remember that H direction is

normal to the plane containing the source. In this

case, square is in xy plane normal to which is a zhence direction of H is a z , as shown in the Fig. 6.8

by right handed screw rule.

� H = 0.4501 a z A/m

� Htotal = 4 H = 1.8 a z A/m

All sides produce H in the same direction at point P.

Example 6.5.7

Solution : The arrangements are shown in the Fig. 6.9



"2

"1

P

r

y

AB

C D

2.5

2.5

x

Fig. 6.7

PComing out ofpaper in directionaz

C

B

!

Fig. 6.8

P

�

y

x

z = –�

P(1,2,3)

y

O

x

"2

"1

r

z

5 m

3 m

(a) (b) (c)

�

P(1,2,3)

y

x

"2

"1

z

5 m

3 m r

Fig. 6.9

Case a : It is infinitely long straight conductor.

H =I

r2 a , P (1, 2, 3), I = 10 A

Now r = x y2 2� � � �1 4 5 m and = tany

xtan 2 63.43º1 1� ��

� H =10

2 5 �a = 0.7117 a A/m

To find x component, take dot product with a x .

� Hx = H a x- = 0.7117 a a x - = – 0.7117 sin

Similarly Hy = H a y- = 0.7117 a a y - = + 0.7117 cos and a a z - � 0

� Hx = – 0.6365, Hy = 0.3183

� H = – 0.6365 a x + 0.3183 a y A/m

Case b : It is a finite length conductor with z 01 � and z2 = 5 m. [Refer Fig. 6.9 (b)]

r = x y 1 4 52 2� � � � m , = tany

x63.43º1� � at point P

"+ = tan �1 3

5= 53.3� but negative as that end is below point P.

� "+ = – 53.3� and ", = tan �1 2

5= 41.81�

H = � �Ir4

" ", + sin sin� a = �� 10

4 5 �� sin 41.81 sin 53.3 a = 0.5225 a

� Hx = H a x- = 0.5225 ( a a x - ) = 0.5225 ��sin

and Hy = H a y- = 0.5225 ( a a y - ) = 0.5225 �cos ... = 63.43º

� Hx = – 0.4673, Hy = 0.2337 i.e. H = – 0.4673 a x + 0.2337 a y A/m

Case c : It is a conductor from z = 5 to z = �. [Refer Fig. 6.9 (c)]

r = x y 1 4 52 2� � � � m, = 63.43º

"+ = tan2r

tan2

541.811 1� �� and ", = tan � �

� �1 90r

Both "+ and ", are positive as above point P.

� H = � �Ir4

" ", + sin sin� a = � �10

4 5 ��sin 90 sin 41.81 a = 0.1186 a

� Hx = H a x- = 0.1186 �a a x - = 0.1186 ��sin

and Hy = H a y- = 0.1186 �a a y - = 0.1186 �cos



� Hx = – 0.106, Hy = 0.053 i.e. H = – 0.106 a x + 0.053 a y A/m

Example 6.5.8

Solution : For a conductor in the form of regular polygon of n sides inscribed in a circleof radius R, the flux density B at the centre is given by,

B =

0 nI

2 R ntan�

��

�� … Refer Ex. 6.5.4

For given conductor, n = 6, R = 1 m, I = 5 A

� B =4 10 6 57

� � �

��

��

�

2 1 6tan = 3.4641 Wb m 2

Example 6.5.9

Solution : Consider the various sections of thecircuits.

Section I : Section AB is shown in the Fig. 6.10 (a),

PM is the perpendicular on AB.

� r = l (PM)

The triangle ABC is right angled triangle.

� x = tan � � �1 43

53.13

� "2 = 90 � � �x 36.8698

From the symmetry of the circuit,

"1 = "2 � �36.8698

But as A is below point P, "1 � � �36.8698

r = l (PM) = BP cos "2 = 25 cos (36.8698�) = 2 m

� H1 = � �I4 r

" "+sin sin2 � a N

= � �104 2

sin 36.86 sin ( 36.86 ) �

�� a N

= 0.4774 a N A/m

Section II : Section B to C along a semicircle as shown in

the Fig. 6.10 (b).

The H at the centre of circular conductor is given by

(I/2R) a N .

� Due to semicircular portion,

H2 =12

I2R

� a N

=10

4 2.5�a N = a N A/m



"2

"1

r=2 mP

B

CA

M

4 m

3 m

2.5

2.55 m

x

Fig. 6.10 (a)

R=2.5B

P

C

Fig. 6.10 (b)

Section III : Section C to A is shown in the Fig. 6.10 (c).

PM is perpendicular on AC.

As triangle ABC is right angled triangle,

x = tan � � �1 34

36.8698

� "1 = 90 – x = 53.13�

From the symmetry of the circuit,

"2 = "1 = 53.13�

But "1 is negative as point C is below point P, "1 = – 53.13�

r = l (PM) = PC cos "1 = 2.5 cos (53.13�) = 1.5

� H3 = � �I4 r

" "+sin sin2 � a N = � �104 1

sin 53.13 sin ( 53.13 ) �

�� .5

a N

= 0.8488 a N A/m

Hence total H at point P is,

H = H H H 2. 3262 a1 2 3 N� � � A/m

Example 6.5.10

Solution : Consider the sections of given loop tocalculate H at P.

Section I : The portion AB of the circuit, as shown in

the Fig. 6.11 (a), PM is the perpendicular on AB from

P. Note that "1 and "2 are to be measured fromperpendicular line from P to the conductor.And r = l (PM)

The triangle ABP is a right angled triangle hence,

x = tan tan� �� 1 1APPB

10.5

63.43 ... from . ABP

� "2 = 90 � � �x 26.565 ... from . PMB

And "1 = 90 2� � �" 63.43

But "1 is negative as point A is below point P.

r = l (PM) = BP cos "2 = 0.5 cos (26.565�) = 0.4472 m

� H1 = � �I4 r




P

2 m 2 m

3 m

4 m

MA C

"2"1

x

2.5

2.5

B

Fig. 6.10 (c)

"1

"2 0.5 m

1 m

x

A

BM

P

r

Fig. 6.11 (a)

= � �54 0.4472

sin 26.565 sin ( 63.43 )�

� � � a N

= 1.1936 a N A/m

where a N = Unit vector normal to the plane is which

the circuit is placed

Section II : The semicircle B to C.

The H at the centre of circular conductor is (I/2R) a N where R is

radius of the conductor.

Hence H2 due to semicircular loop is,

H2 =12

I2R

a5

4 0.5a 2.5 aN N N�

�� A/m

Section III : The portion C to A is shown in the Fig. 6.11 (c).

PM is perpendicular on AC.

x = tan � � �1 0.51

26.565 as triangle

APC is right angled triangle

� "2 = 90 – x = 63.43�

And "1 = 90 – "2 = 26.565�But "1 is negative as point C is below point P.

r = l (PM) = PC cos "1 = 0.5 cos (26.565�) = 0.4472 m

� H3 = � �I4 r


= � �54 0.4472

sin 63.43 sin ( 26.565 )�

�� a N � "1 26 56� – . º

= 1.1936 a N A/mHence the total H at point P is

H = � �H H H a1 2 3 N� � � � �1.1936 2.5 1.1936

= 4.8873 a N A/m

Example 6.5.11

Solution : Consider the arrangement as shown in theFig. 6.12.

"1 0� and "2 = � �tan 1 25

= –21.801�

H =I

4 r[sin sin ]" "2 1� a N



!R=0.5 m

B

C

P

Fig. 6.11 (b)

90º

"1

"2 0.5 m

1 m

xA

CM

P

r

Fig. 6.11 (c)

Side 1

r = 5

! = 10 A

"2

P(0, 0, 5)

x

z

0 2

Fig. 6.12

where a N = a y"1 and "2 are negative as both the ends of the conductors are below point P.

� H =10

4 5�� [sin( 21.801 ) sin(0 )] ( )a y = – 0.0591 a y A/m

Example 6.7.6

Solution : Kept this unsolved example for students' practice.

Example 6.7.7

Solution : Refer the procedure discussed in the section 6.7.

H =Ir

2 (r z )

2

2 2 3/ 2�a z A/m

In this example, r = 5 cm and I = 50 mA. The ring is in z = 1 plane and point P (0, 0, – 1)

hence z = 2 cm.

� H =50 10 5 10

2 5 10 2 10

3 2 2

2 2 2 2 3 2

� � �

� � �

� �

� �

( )

[( ) ( ) ] /a z = 0.4 a z A/m

Example 6.9.9

Solution : The planes are shown in the Fig. 6.13.

For sheet in z = 0 plane,

a N = a z … towards P

For sheet in z = 4 plane,

a N = – a z … towards P

H1 =12

K a N� =12

8� � �( )a ax z = 4 a y

H2 =12

K a N� =12

18� ��( )a ax z = 9 a y

� H at P = H1 + H2 = 13 a y A/m.

Example 6.9.10

Solution : The current from current density is given by,

I = � /J Sd

dS = r dr d a z normal to a z as J is in a z

� I =

� 0

,

� ��

� /r 0

r2r4.5 e r dr da az z = 4.5

� 0

,

� ��

�

r 0

r2rr e dr d

Using integration by parts,

= 4.5 ) * � 0

,

� � � �� d r e dr 1 e dr dr2r 2r

0

r= 4.5 �2

r e2

e2

dr2r 2r

0

r� �

��

�

#$1

%1

&'1

(1�



y

x

z

z = 4

z = 0

K = – 8 ax

P (1,1,1)

K = 18 ax

Fig. 6.13

= 9 r e

212

e2

2r 2r

0

r� �

��

�

#$1

%1

&'1

(1= 9

9

4

��

#$1

%1

&'1

(1� � �

�� r e

214

e14

1 2 r e e2r

2r 2r 2) *r A

For r = 0.5, I = 7.068 [1 – 0.3678 – 0.3678] = 1.8676 A

Consider a closed path with r 0.52 such that the enclosed current I is 1.8676 A.

According to Ampere's circuital law,

� /H Ld = I

� � 0

,

� H r d = I ... H = H d a L, = r d a

� 2 r H = 1.8676 i.e. H =1.8676

r0.2972

r2 �

� H =0.2972

ra A/m for r 0.52 m

Example 6.9.11

Solution : For a given point P (0.01, 0, 0).

r = ( . ) ( )0 01 02 2� = 0.01 m = 10 mm

Thus P is in the region b < r < c.

� H =I

2 rc r

c b

2 2

2 2 �

�

�

��

�

��

a =6

2 0.01

(0.011) (0.01)

(0.011) (0.009)

2 2

2 2��

�

�

��

�

��

= 50.113 a A/m

Example 6.9.12

Solution : Due to infinite long conductor along z-axis.

H1 =I

2 r 2 r a a�� 20 10 3

=10 10 3� �

ra A/m

At r = 0.5 cm, no current sheet is enclosed.

� H = H a 2 a1 ��

��

�

�

10 10 3

0.5 10 2 A/m

At r = 1.5 cm, current sheet at r 11 � cm is getting enclosed. It carries current in z

direction.

� K1 = 400 � �10 3 a z A/m

I enc = K 2 r1 1� � � � � �� 400 10 2 1 103 2 = 0.02513 A ... r1 = 1 cm for sheet


� /H Ld = I enc i.e.

= 0

2

H r d� = I enc ... H = H a and dL = r d a



� H (2 r) = I enc i.e. H =I2 r

0.025132 r

enc

�

� H2 =4 10

r

3� �a A/m and H1 =

1 10r

30� �a A/m

So at r = 1.5 cm, H = H H a 0.933 a1 2� ��

��

�

�

14 10

1.5

3

10 2 A/m

At r = 2.5 cm, second sheet also gets enclosed for which,

K2 = � � �250 10 3 a z A/m

� I enc = K 2 r2 2� � � � � � �� 250 10 2 2 103 2 = – 0.03141 A ... r2 = 2 cm for sheet


� /H Ld = I enc i.e.

= 0

2

H r d� = I enc

� H =I2 r

0.03141

2 renc

��

i.e. H3 =� � �5 10

r

3

a A/m

So at r = 2.5 cm, H = H H H a1 2 3� � � � ��

�� 4 10 5

10 3r r r

=9r

10

2.5 10

3

2� �

�

��

�

�10

93 a a 0.36 a A/m

Example 6.9.13

Solution : The planes are shown in the Fig. 6.14.

i) P(1, 1, 1)

For z = 0 plane, a aN z� � at P

� H1 = �� 12

12

–10K a a aN x z� ��

= � �–102

– 5a ay y� � A/m

For z = 4 plane, a aN z� – at P as P is below

z = 4 plane.

� H2 = � �� 12

12

–10K a a –aN x z� ��

= �� 102

– 5– a ay y� � A/m

� H = H H1 2� = 10 a y A/m at P(1, 1, 1)

ii) Q(0, –3, 10)

For z = 0 plane, a aN z� � at Q hence H a1 y� �5 A/m.



z

y

x K a= –10 x z = 0

Q

(0,–3,10) K a= +10 x z = 4

P(1,1,1)

Fig. 6.14

For z = 4 plane, a aN z� � at Q as Q is above the plane

� H2 = �� 12

10K a a aN x z� ��12

� a a –ax z y� �

= � �102

– –5a ay y� A/m

� H = H H1 2� = 0 A/m at Q (0, –3, 10)

Example 6.9.14

Solution : The sheet is located at y = 1 on

which K is in a z direction. The sheet is infinite

and is shown in the Fig. 6.15.

The H will be in x direction.

a) Point A �0 0 0, ,

a N = �a y normal to current sheet at point A

� H =12

K a N�

= � �12

40a az y��

Now a az y� = �a x

� H = � �12

40� �a 20 ax x A/m

b) Point B �1 5 2, , �

This is to the right of the plane as y = 5 for B.

� a N = a y normal to sheet at point B

� H = � �12

12

40K a a a 20 aN z y x� �� A/m

Example 6.10.6

Solution : H = � � � �y(x y ) x(x y )2 2 2 2a ax y , H 0z �

� Hx = � � �� y(x y ) yx y2 2 2 3 , H x(x y ) x xyy2 2 3 2� � � �

J = .�H =

a a ax y z33

33

33x y z

H H Hx y z

= a ax yyH

zH

xH

yHz y z x

33

33

33

33

��

��

��

��

�� xH

yHy x� ��

��

��a z

33

33

33x

Hy =33x

[x xy ] 3x y3 2 2 2� � �



y

x

z

K a= 40 z

y = 1plane

Fig. 6.15

33y

Hx =33y

[ yx y ] x 3y2 3 2 2� � � � �

� J = a az z[3x y (x 3y )] [4x 4y ] A m2 2 2 2 2 2 2� � � � �

I =

S

. d� J S where dS = dx dy a z

� I =

S

2 2[4x 4y ] dx dy.� � a az z

=

y x� � � � � ��

2

2

1

1

4[ x 4y ] dx dy4x

34y x2 2

y 2

2 32�

��

�

�� x 1

1

dy

=

y 2

22 2

y 2

43

4y43

4y dy

� � � ��

��

�2

283

8y dy� ��

��

=83

y8y

3163

163

643

y 2

2

��

��

�

��

� � ��

33643

� � 53.333 A

This can be cross checked by calculating current through the given region –1 < x < 1 and

–2 < y < 2 by using I = � H L. d .

Consider the region in z = 0 plane as shown in the

Fig. 6.16.

I = � H L. d

=

A

B

B

C

C

D

D

A

d.� � � �� H L

For A - B, dL = dx a x , H L.d = � �y(x y )dx2 2 , y = –2

For B - C, dL = dy a y , H L.d = x(x y )dy2 2� , x = 1

For C - D, dL = dx a x , H L.d = –y(x y )dy2 2� , y = + 2

For D - A, dL = dy a y , H L.d = x(x y )dy2 2� , x = – 1

� I =

x 1

12

y 2

22( 2) [x 4] dx (1)[1 y ] dy

� � � ��

�

�

�

�

� �2 (x 4) dx ( 1) (1 y ) dy

x 1

12

y 2

22

= 2x3

4x yy

32

x3

3

1

1 3

2

23

��

��

�

��

� ��

��

�

��

� ��

4x yy

31

1 3

2

2�

��

�

��

� ��

��

�

��

� �



– 1

– 2

2

1x

y

A B

CD

Fig. 6.16

= 213

4+13

4 2+83

283

2� ��

��

��

13

413

4 283

283

� � ��

��

��

= (2 8.66) 9.333 (2 8.66) 9.333� � � � � �53.33 A

Thus, � �� H L J S. .d d I

S

Example 6.10.7

Solution : i) H a a Jy z� x z – y x ,2 2 = 4�H

J =

a a ax y z33

33

33

0

x y z

x z –y x2 2

= –2xy 2xz y – x2 2a a a ax z y x� �

At PA (4, 3, 4), J = – 40 a 9 a 32 ax y z� � A/m 2

ii) H =25

(cos 0.2 ) a , H5= Hz = 0

J = 4�H =1

zH H H

1z

z

z

5

5335

33

33

55

5

335

33

33

5

5

5 a a a a a az

�

0 2 0cos .2 0

= 0 A/m 2

iii) H =1

sin � �a , Hr = H = 0, H� =1

sin �

J = 4�H =1

r sin

r r sin

rH r H r sin H

2

r�

�33

33�

33 �

�

�

a a ar

�1

r sin

r r sin

r

0r

sin0

2 �

�33

33�

33

�

� a a ar

=1

r sin � a

At PC(2, 30º, 20º), J =1

2 sin 30º�a = a A/m 2 .

Example 6.10.8

Solution : From the point form of Ampere's circuital law,

4�H = J

In the cartesian system,

4�H =

�

a a ax y z33

33

33�

x y z

y x y excos " 0



= � � �33

��

��

��

33

�3 �

3

�

�

��

�

�

��y

y ey x

z

y e

xx

x

a ax ycos "

+ ��33

�

��

��yy xcos " a z

= � � �1 a a ax y z� � � �0 e xx cos"

On yz plane, x = 0

� J on yz plane = a a ax y z� �e0 cos 0 = a a a A / mx y z2� �

Example 6.10.9

Solution : In cylindrical co-ordinates 4�A is given by,

4�A = �1

r

A A

z

A

z

A

r1r

r Az r z3

3�3

3�

��

�

��

33

�33

�

��

��

3

3

a ar r1r

A r�33

�

�

��

�

�

��

a z

Now A r = 0, A = sin 2 and A z = 0

� 4�A = � � �

02

0 02

0�33

�

��

��

33

��

��

�

��

sin sin z

1r

r

ra a ar z

= � �0 0 02 2

� � � �a a a ar z z sin sin

r r

At 24

0, ,�

��

�� , r = 2, �

�

, z = 0

� 4�A =

sin 24

2

��

��

a z =

sin,

,

��

��

a z = 0.5 a z

Example 6.10.10

Solution : In the spherical co-ordinates, curl H is given by,

4�H = �1 1

r

H H 1r

H r H

rr

sin

sin

sin�

�

� � � 3

3�33

�

��

�

��

33

�3

3a r

�

�

��

�

�

��

a �

��

33

�33

�

��

�

��

1r

r H

r

Hr� �

a

Now Hr = 0, H� = 2.5, H = 5

� 4�H = �1 5

0r

2.5 1r

5 r

rsin

sin

��

� �3

3�33

�

��

��

33

�

��

�

��a ar +

�1r

r 2.5

r

33

��

��

�

��0 a

= � � � � � �1

5 5r

1r

1r

2.5sin

cos�

��0 � a a ar � � �

=5r

5r

2.5r

cot � � a a ar � �



At 26

0, ,�

��

�� , r = 2, � �

6

, � �0

� 4�H =52

52

cot6 � a a ar � �

2.52

= 4.33 a r � 2.5 a � + 1.25 a

Example 6.11.3

Solution : According to Stoke's theorem,

L

d� /H L =

S

d� 4 � /( )H S

Let us evaluate left hand side. The

integral to be evaluated on a

perimeter of a closed path shown in

the Fig. 6.17. The direction is

a-b-c-d-a such that normal to it is

positive a z according to right hand

rule.

� /H Ld =

ab bc cd da

d� � � �� /H L

ab

d� /H L =

x 2

526xy 3 y dx

�� /( )a a ax y x

=

x 2

5 2

2

5

6 xy dx 6yx2

��

�

��

�

��

=6y

225 4][ � = 63 y

Now y = – 1 for path ab,

ab

d� / � � � �H L 63 1 63( )

Similarly

bc

d� /H L =

y 1

12

33

113y dy

3 y

3y ] 1 1

� ��

�� [ [ ( )] = – 2

cd

d� /H L = �x 5

2 2

5

2

6 xy dx 6x2

y6y

24 25]

��

�

��

�

��

� � �[ � 63y

But y = 1 for path cd, hence

cd

d� /H L = – 63

da

d� /H L =

y 1

12 3

11 3 33y dy y ] 1 1 ] 1 1]

�

�� [ [( ) ( ) [ 2

� � /H Ld = � � � � � �63 2 63 2 126 A

Now evaluate right hand side.



a

bc

d x = 2

x = 5

x

y = –1y = 1

y

z

Fig. 6.17

4�H =

a a ax y z33

33

33

�

x y z

xy 3y 26 0

= a a a a ax y z z z[ ] [ ] [ ]0 0 0 0� � � � � � �0 6x 6x

�S

d� 4 � /( )H S =

S

6 x dx dy� � /( ) ( )a az z

dS = dx dy a z normal to direction a z

� �S

d� 4 � /H S =

y 1

1

x 2

5 2

2

5

11

6x dx dy 6x2

y]

��

�

��

�

��

[

= � � � � � � � � � �62

25 4 1 1 3 21 2[ ][ ( )] 126 A

Thus both the sides are same, hence Stoke's theorem is verified.

Example 6.11.4

Solution : According to Stoke’s theorem,

L� /H dL = �

S

d� 4 � /H S

In spherical system,

dL = dr a a ar � �r d r sin d� � �

The closed path forming itsperimeter is composed of 3paths as shown in theFig. 6.18.

For all the three paths r = 4 m,

For arc 1, r = 4, � �0 and � ischanged from 0 to 0.1 .

For arc 2, r = 4, � = 0.1 and is changed from 0 to 0.3 .

For arc 3, r = 4, = 0.3 and� is changed from 0.1 to 0 tocomplete the closed path.



Path 2

d = r sin dL a�

0.3

= 0º

0.1

� = 0.1 , r = 4

Path1r = 4

= 0º

d = rd

L a��

Path3 r = 4,

= 0.3

d = rd

L a��r = 4

, = 0.3

Fig. 6.18

�L� /H dL =

path 1 path 2 path 3� � �/ / /� �H dL H dL H dL

Now Hr = 6r sin , H� = 0, H = 18 r sin � cos

For path 1, dL = r d� �a and H� = 0 so

path 1� /H dL = 0

For path 2, dL = r sin � d a and H = 18 r sin � cos

�path 2� /H dL =

� �0

0.32 2 2 218 r sin cos d 18 r sin sin

��

00.3

= 18 r sin 0.80902 2�� = 22.2487 ... r = 4, � = 0.1

For path 3, dL = r d� a � and H� = 0 so

path 3� /H dL = 0

� � /H dL = 22.2487 A

Let us find �S

d� 4 � /H S

4�H =1

r sin

H sin H 1r

1sin�

�

� � �

3

3�33

�

��

�

��

3a r

�H d r H

rr

3�

3

�

�

��

�

�

��

�a

��

33

�33

�

��

�

��

1r

r H

rHr�

�a

= � �1

r sin36 r cos sin cos 0

1r

1sin

6r�

� ��

� �a r cos 36 r sin cos � ��

��

a + � �1r

0

Now dS = r sin d d2 � � a r ... In � a r direction

� �4� /H Sd = � �36 cos cos r sin d d2� � � ... r = 4

� �S

d� 4 � /H S =

7

�

7

� � � 0

0 3

0

0 1236 r sin cos cos d d

� ��

= 18 r cos d sin 2 d2

0

0 3

0

0.1

7

�

� �� ... 2 sin � cos � = sin 2 �

= � �18 r sincos 2

22

00.3

0

0.1

�

��

��

... r = 4

= � �18 16 0.809012

cos 36 1� � � � �� = 22.2487 A

Thus Stoke’s theorem is verified.



Example 6.12.4

Solution : The surface is shown in the Fig. 6.19.

The flux crossing the surface is given by, =

S

d� -B S

The dS normal to a direction is dr dz.

� dS = dr dz a

� =

S

2.0r

dr dz� -a a

=

z 0

2

r 0.5

2.52.0r

dr dz

� ��

= 2.0 � � � �ln r z0.52.5

02

= 2.0 [ln 2.5 – ln 0.5] [2 – 0] = 6.4377 Wb

Example 6.12.5


B =

0

6 7

r2.239 10 4 10

rcosH a�

� � � �

=2.8136

rcos Wb mr

2 a

=

s

d� -B S

= v

4

4

z 0

12.8136

rcos r d dz

� � ��

= 2.8136 [sin ] [z]4

401

� �

= 2.8136 [0.7071 ( 0.7071)]� � � � 3.9789 Wb

Example 6.13.6

Solution : Vector magnetic potential, A a z�50 r 2 Wb/m

Now, B = 4 �A

=1r

A A

zA

zA

rz r z3

3�3

3�

��

�

��

33

�33

��

��

a ar �

3

3�33

�

��

�

��

1r

rA

rA r

( )

a z

Now A 0, A 0, A 50 rr z2� � �



r = 0.5

z = 2.0

r = 2.5

O

z

a

in directiona

Fig. 6.19

z

dS

– /4 /4

Fig. 6.20

� B =1r

r ) (50 r

r

2 233

��

��

�

��

� �33

�

��

�

��

( )500 0

a ar � �� 1r

0 0 a z = �100 r a Wb/m2

� H =B

a

0 0

100r�

�A/m

Now J = 4 � H

Hr = 0, H100 r

, H 00

z � � �

� 4 � H = � �0 0 0�

3 ��

��

�

��

3

�

�

��

�

�

��

� � �

3�100 r

z1r

100 r

0 a ar

2

0

r

�

��

�

��

3�

�

�

��

�

�

��

0 a z

= � � � �0 0 0100

� � � ��

��

��a a ar z

1r

2r0

= �200 0

a z A/m2

� J = �200 0

a z

Now I =

S

d� /J S where d = r dr dS a z

=

� � � /= 0

2

r = 0

1

0r dr d

200a az z =

� � �= 0

2

r = 0

1

0r dr d

200

= � � � ��

��

�

��

��

��

200 200 12

2

0

1

0

2

02

0

r2

= � �500 106 A

So current is 500 MA and negative sign indicates the direction of current.

Example 6.13.7

Solution : The straight conductor of

length '2L' carrying current I is placed

along z axis as shown in the Fig. 6.21.

Consider the differential current element

along the conductor, at the origin.

� IdL = I dz a zThe vector magnetic potential A due to

differential current element is given by,

� A = �

0 I d

4 R

L



O

P(x,y,z)

y

r

! dL

!

+L

–L

x

R

z

y

x

x

y

z

Fig. 6.21

R = � � �x 0 y z 02 2 2� � � � �0

= �x y z2 2 2� �1 2/

� A =

��

� �L

+ L0

2 2 2

I dz

4 x y z

1 2/

a z

Converting P (x, y, z) to cylindrical co-ordinates,

r = x y i. e. x y r2 2 2 2 2� � �

� A =

�z= L

+ L0

2 2

I dz

4 r z��

�

a z1 2/

=

�2

01 2

z=

L0

2 2

I dz

4 r z�

�

/

a z

Now ��

dx

(x a )2 2 1/ 2= ln x x a2 2� ��

��

�� hence

A =

0 2 2

LI

2z z rln � ��

��

��

��

�� 0

a z

=

0 2 2 2 2I

2L L r 0 0 rln ln� ��

��

��

��

��

��

��

a z

� A =

0

2 2I

2L L r

rln

� ��

�

��

�

�

��

a z Wb/m ... Cylindrical

� A =

0

2 2 2

2 2

I

2

L L x y

x + yln

� � ��

�

��

�

�

��

a z Wb/m ... Cartesian

For L to be long, L >> r hence L r L2 2 2� 8

� A =

0

20I

2L L

r

I

22Lr

ln ln��

�

��

�

�

��

� ��

��a az z Wb/m

Now B = 4 �A

A 0r � , A = 0 AI

22Lrz

0� ��

��

ln

� 4 �A =1r

A A

zA

zA

rz r z3

3�3

3�

��

�

��

33

�33

��

��

�

a ar

1r

rA

rA r

3

3�33

�

��

�

��

( )

a z

= � �0 0a a ar z�33

�r

A z



� B = �33

��

��

�

��

��

0 I

2 r2Lr

ln a

= � � ��

��

�33

��

��

0 I

2 2Lr

r2Lr

ddx

u1u

dud

1a ... ln

x

= � � ��

��

�

��

02

I

2rL

r L) (1)

r2

0 2( ) (a = � � �

�

02

I

2rL

2L

r2a

� B = 0 I

2 ra Wb/m2

� H =B

a

0

I2 r

� A/m

��




Example 7.2.6

Solution : The magnitude of velocity is given as v � �6 106 m/s. The direction of this

velocity is specified by an unit vector. Thus we can write,

v = v 0.48 0.6 0.64va a a ax y z� � � � �6 106 [ ] m/s

The force experience by a moving charge in a steady magnetic field B is given by,

F = Q v B�

= � � � � � � � �� 60 10 6 10 2 6 59 6[( ) ( ) (0.48 0.6 0.64a a a a a ax y z x y z ) ( )]1 10 3� �

= ( )� � � �

�

�3.6 10 0.48 0.6 0.644

a a ax y z

2 6 5

= ( ) [ ]� � � ��3.6 10 0.84 3.68 4.084 a a ax y z

= ( )� � � � �0.3024 1.3248 1.4688a a ax y z 10 3 N

Thus the magnitude of the force on a moving charge is given by,

F = ( ) ( ) ( )� � � � � � � �� 0.3024 10 1.3248 10 1.4688 103 2 3 2 3 2 = 2.0009 mN

Example 7.2.7

Solution : Given : Q = – 40 nC = – 40 � 10–9

v = 6 10 = 6 10 (– 0.48 – 0.6 + 0.64 )6 6� �a a a av x y z m/s

i) The magnetic force exerted by B on charge Q is given by,

Fm = Q v � B

= – 40 10 [6 10 (– 0.48 – 0.6 + 0.64 ) (2–9 6z� � �a a ax y – 3 + 5 ) 10–3a a ax y z � ]

= – . .

–

–2 4 10 0 64

2 3 5

4�

a a ax y z

– 0.48 – 0.6


7 Magnetic Forces and Inductance

(7 - 1)

= – 2.4 10 [– 1.08 a 3.68 a 2.64 a ]– 4x y z� � �

= [0. 2592 a 0.8832 a 0.6336 a ] mNx y z– –

Hence the magnitude of the force is given by,

| |mF = (0.2592 10 ) + (– 0.8832 10 ) + (– 0.6336 10 )–3 2 3 2 3 2� � �– –

� | |mF = 1.1174 10 N = 1.1174 mN–3�

ii) The electric force exerted by E on charge Q is given by,

Fe = Q � E

= – 40� �10 (2 – 3 + 5 ) 10–9 3a a ax y z

= (– 80 + 120 – 200 ) 10 N– 6a a ax y z �

The magnitude of the electric force is given by,

| |eF = (– 80 10 ) + (120 10 ) + (– 200 10 )– 6 2 6 2 6 2� � �– – = 0.2465 mN

iii) The total force exerted by both the fields B and E together is given by,

F = Fm + Fe = (0.2592 a a ax y z– 0.8832 – 0.6336 ) 10– 3�

+ (– 80 + 120 – 200 ) 10x y z– 6a a a �

= [( . – . ) –0 (– 0.8832+ 0.12) 103 – 32592 0 08 10� � �a ax y

+ (– 0.6336 – 0.2) 10– 3� a z

� F = 0.1792 10 – 0.7632 10 – 0.8336 10– 3 – 3 – 3� � �a a az y z

� F = (0.1792 – 0.7632 – 0.8336 ) mNa a ax y z

Hence magnitude of the total force is given by,

| |F = (0.1792 10 ) + (– 0.7632 10 ) + (– 0.8336 10 )– 3 2 – 3 2 – 3 2� � �

� | |F = 1.1443 10 N– 3� = 1.1443 mN

Example 7.2.8

Solution : i) Let the position of the charge is given by P(x, y, z).

The force exerted on charge by E is given by,

F = QE …(1)

According to Newton's second law,

F = m a = mddt

d

dt

2

2

v z� …(2)

Electromagnetic Theory and Transmission Lines 7 - 2 Magnetic Forces and Inductance


Equating equations (1) and (2) we can write,

md

dt

2

2

z= Q �E = (– 0.3 10 30 )– 6� � a z …(3)

The initial velocity is constant and it is in x-direction so no force is applied in that

direction. Rewritting equation (3), we get,

d

dt

2

2

z=

Q

m

E…(4)

Integrating once equation (4) by separating variables, we get

ddtz

= vE

z =Q

mt k1� …(5)

Where k1 is constant of integration.

To find k1 : At t = 0, initial velocity in z-direction is zero Substituting values in

equation (5), we get

0 = 0 + k1 i.e. k1 = 0

Thus equation (5) becomes,

vz

�ddt

=Q

mt

E…(6)

Integrating equation (6) with respect to corresponding variables we get,

z =Q

mt2

k2

2E

��

�

�� …(7)

Where k2 constant of integration.

To find k2 : At t = 0, charge is at origin. Substituting values in equation (7) we get,

0 =Q

m02

k2E �

� � � i.e. k2 = 0

Hence solution of the equation (3) is given by,

z =Q

2mt =

– 0.3 10t2

– 6z 2E a� �

� �

30

2 3 10 16–…(8)

At t = 3 �sec,

z =– 0.3 10– 6� �

� ��

30

2 3 103 10

166 2

––( ) = – 0.135 m

Let us consider initial constant velocity in x-direction, the charge attains x-co-ordinate of,

x = vt = ( )( )–3 10 3 105 6� � = 0.9 m

Hence at t = 3 �sec, the position of charge is given by,

P(x, y, z) = (0.9, 0, – 0.135) mii) To find velocity at t = 3 � sec using equation (6), we get,

v =Q

mt =

– 0.3 10 30 )

3 10

– 6z

– 16

E a� �

��( )–3 10 6

= – 9 � 104 a z m/sec



The actual velocity of charge can be obtained by including initial constant velocity in

x-direction as,

v = (3 10 a – 9 10 a ) m / sec5x

4z� �

iii) The kinetic energy of the charge is given by,

K.E. =12

m | |2v � � � � � � �12

3 10 3 10 9 1016 5 2 4 2 2– [ ( ) (– ) ]

� K.E. = 1. 4715 10 J– 5�

Example 7.2.9

Solution : Q = Charge = 5 10 C– 18�

B = – 0.4 + 0.2 – 0.1 Ta a ax y z

v = (2 – 3 + 6 ) 10 m / s5a a ax y z �

(a) By definition,

F = Q ( + )E v B�

At t = 0, F = 0

� 0 = 5 10 [ + (2 – 3 + 6 )10 (– 0.4 + 0.2 – 0.1– 18 5� �E a a a a ax y z x y a z )]

� E = – [(2 – 3 + 6 )10 (– 0.4 + 0.2 – 0.1 )]5a a a a a ax y z x y z�

� E = – 105

a a ax y z

2 – 3 6

– 0.4 0.2 – 0.1

� E = – 10 [(+0.3 – 1.2) – (– 0.2+ 2.4) + (0.4 – 1.2) ]5 a a ax y z

� E = 0.9 10 + 2.2 10 + 0.8 105 5 5� � �a a ax y z

� E = (0.9 a + 2.2 a + 0.8 a )10 V / mx y z5

(b) Let E field be in x-direction only. Then we can write,

E = Ex a x

The force is given by,

F = Q ( + + )E v B

� F = 5� �10 [(E )+ {(2 – 3 + 6 )10 (– 0.4 + 0.2– 18x

5a a a a ax x y z x a ay z– 0.1 }]

� F = 5� � � �10 [(E – (0.9 10 + 2.2 10 + 0.8 10 )]– 18x

5 5 5a a a ax x y z

� F = 5 2 8�10 [(E – 90000) – 2000 – 0000 ]– 18x a a ax y z



But| | = 2 10–12F � N. Hence finding magnitudes on both the sides and equating,

� 2 10– 12� = 5 10 (E – 90000) + (– 220000) + (– 8– 18x

2 2� 0000 2)

� (E – 90000) + (– 220000) + (– 80000) = (400000)x2 2 2 2

Solving for Ex , we get

� (E – 90000)x2 = 1.052 1011�

� Taking square root we get,

Ex – 90000 = � 324345.5

� Ex = 414. 355 10 V / m or E = – 234.3955 V / m3x�

Example 7.3.7

Solution : A force exerted on current carrying conductor in a magnetic field is given by

F = I dL B� = � � � �2 6a az x� 0.08 = 12 a az x� 0.08

� F = 0.96 a y N … a a az x y� �

Example 7.3.8

Solution : In the plane z = 0, the z-component of the magnetic field B will be zero. In

other words it will not contribute to the force. The force on the loop is given by,

F = � �I dL B

Considering dimensions of all sides in meter, we can write,

F =

x = 0.01

x = 0.03

y = 0.02

y

30 dx (– 3x + 5y )+� �a a ax x y

= 0.05

30 dy (– 3x + 5y )� �a a ay x y

+

x = 0.03

x = 0.01

y = 0.05

y

30 dx (– 3x + 5y )+� �a a ax x y

= 0.02

30 dy (– 3x + 5y )� �a a ay x y

For side 1 and 3, the values of y are 0.02 and 0.05 respectively. While for sides 2 and 4,

the values of x are 0.03 and 0.01 respectively.

� F =

0.01

0.03

0.02

0.05

30 dx (5) (0.02) 30 dy (–3) (� ��a z 0.03) (– a z )

+

0.03

0.01

0.05

0.02

30 dx (5) (0.05) 30 dy (–3) (� ��a z 0.01) (– a z )

� F = 3 7 5

0 03

0 01

0.01

0.03

0.02

0.05

dx 2.7 dy dx� � �� a a az z z.

.

.

+ 0.9 dy

0.05

0.02

� a z

� F = � � � � � �3 x + 2.7 y + 7.5 x + 0.0 01

0 03

0 020 05

0 03

0 01

.

.

.

.

.

.a a az y z � �9 y

0 050 02.. a z



� F = [3(0.03 – 0.01)+ 2.7(0.05 – 0.02)+ 7.5(0.01 – 0.03)+ 0.9(0.02 – 0.05)]a z

� F = – 0.036 a = – 36 a mNz z

Example 7.3.9

Solution : The magnetic field intensity due to straight filament is given by,

H =I

x2�a z =

152�x

a z A/m

But B = � �0 r H = � 0 H = 4 10 7�� 152�x

a z =30 10 7� �

xa z =

3 10 6� �

xa z T

Now force exerted on loop is given by,

F = – I dB L� �

Assuming that the total force on the loop is sum of all the forces on four sides.

F = – 2 10

3 10 3 10

3 1

3 1

3 6 6

0

2

�

��

��

��

� �

� �

��

xx

dx3

dya a a az x z yy

0 3 106 6

2

0

3

1 � �

��

� ��

�

�

�

��

�

�

��

�� xdx

1dya a a az x z y

yx��

� F = � � � �– 2 10 3 103 1

3 6

0

2

2

0

� � � � � � � ��

� ��

dxx

dy dxx

dy

y yx

a a a ay x y x��

�

�

��

�

�

��

3

1

1

3

x

� F = � � � � � � � � � � � ��

��

�6 1013

913

02

30

20

ln lnx y x ya a a ay x y x

� F = � � � ��

��

�6 10 3 113

2 0 0 3 0 29 ln ln ln lna a a ay x y x

� F = � � �8 10 a12x N = � 8 a pNx

Example 7.3.10

Solution : The force exerted on a straight conductor is given by,

F = I L B� = 5(– 0. 3 ) (3.5 10 )– 2a ax z� �

= – 0.0525 (– )a y …( a a –a )x z y� � �

= – 0. 0525(– )a y = 0. 0525 Na y

Hence the force applied to hold the conductor in position must be

F = – 0. 0525 a Ny

Example 7.3.11

Solution : A force exerted on a current element in a magnetic field is given by,

F = I dL B�But current element is 4 cm long i.e. 0.04 m long. It carriers current of 10 mA in

y-direction. The magnetic field is given by,



H =5�

a x A/m

But B = � H = ��

a x��

��

= 5 a x T.

Hence, force exerted is given by,

F = {10 10 [0.04 ]} {53� � �a ay x }

� F = � � �2 10 a N3z = �2 a mNz s

Example 7.3.12

Solution : Diameter of identical loop = 1 m. Hence radius =1 m

20.5 m�

Length of identical loops = l l1 2� = Circumference of a circular loop = 2 r� = � m

Assume that two loops are placed so close to each other that the distance of separation i.e.

d between two identical circular loops is very small compared with diameters of the loops.

Hence assuming that the magnetic field near to the circular loops is approximately equal

to that near to a straight conductors. Let us assume that distance of separation between

two loops is d = 0.1 m. Then by formula the force between two current carrying

conductors is given by,

F =��

� ��

I I

2 d

(100) (100) ( )

2 (0.1)1 2 0l

� ... (� � 0� for free space)

=4 10 100 100

0.2

7� ��

� � � ��

�= 0.06283 N

If the currents in two conducting loops are with same orientations (i.e. in same direction),

then the circular loops will experience force of attraction. While if the directions of

currents in the circular conducting loops are different, then the circular loops will

experience force of repulsion.

Example 7.4.5

Solution : I 1 = I 2 = 100 A

d = Distance of separation = 2 cm = 2 10 2� � m

� � � � � �0 0r .... for air � r = 1

The force per meter length between two long current carrying conductors is given by,

FL

=�

�I I

2 d1 2 =

4 10 100

2 2 10

7 2

2

� � �

� � �

�

�

�

�

( )= 0.1 N/m

As the currents in parallel conductors are flowing in opposite direction, the force will be

force of repulsion.

Example 7.4.6

Solution : The magnetic field intensity at point P1 due to I 1 L1 can be obtained using

Biot-Savart's law as follows.



dH1 =� �

I

R

1

122

L a1 R12�

4 �

The unit vector in the direction of R12 is drawn from P1 to P2 .

� R12 = ( )� � � �1 1 2a ax x

R12 = R12 � �( )2 2 = 2

� a R12 =R

R

a2

a12

12

xx�

��

2

Substituting values of I 1 L a1 R12, and R12 in the expression for dH1 , we get,

dH1 =( ) ( )

( )

10

4

5

2

� � �a az x

� != �

�1016

5

�a y A/m

Thus the magnetic flux density at point P1 is given by,

dB1 = � 0 dH1 = � �4 1010

167

5

��

��

��

�

��

��

a y = – 0.25 � �10 12 a y T

Now the force exerted on I 2 L2 due to I 1 L1 is given by,

F2 = I d2 L B2 1� = 10 2 35� �� ( ) ( )0.6 0.25 10 12a a a ax y z y

= 10 2 317� � � ��[( ) ( )]0.6 0.25a a a ax y z y

= [ ] ( )� � �0.15 0.75a az x 10 17 = ( )7.5 a 1.5 a 10x z18� � N

Example 7.4.7

Solution : I 1 = I 2 = 40 A, � r 1� ....... For air

d = Distance of separation = 5 cm = 5 10 m2� �

The force per meter length L between two long conductors is given by,

FL

=�

��

��I I

2 d

I I

2 d4 10 11 2 0 r 1 2

7� �

� � � �� 40 40

2 5 10 2

�

� � � ��= 6.4 10 N m3� �

As the currents in both the parallel conductors flow in same direction, the forceexperienced will be force of attraction.

Example 7.4.8

Solution : Diameter of identical loop = 1 m. Hence radius =1 m

20.5 m�

Length of identical loops = l l1 2� = Circumference of a circular loop = 2 r� = � m

Assume that two loops are placed so close to each other that the distance of separation i.e.

d between two identical circular loops is very small compared with diameters of the loops.

Hence assuming that the magnetic field near to the circular loops is approximately equal

to that near to a straight conductors. Let us assume that distance of separation between



two loops is d = 0.1 m. Then by formula the force between two current carrying

conductors is given by,

F =��

� ��

I I

2 d

(100) (100) ( )

2 (0.1)1 2 0l

� ... (� � 0� for free space)

=4 10 100 100

0.2

7� ��

� � � ��

�= 0.06283 N

If the currents in two conducting loops are with same orientations (i.e. in same direction),

then the circular loops will experience force of attraction. While if the directions of

currents in the circular conducting loops are different, then the circular loops will

experience force of repulsion.

Example 7.5.5

Solution : Given

I = 100 mA = 100 10 A3� �

B = 0.2 0.1 0.2 Ta a ax y z� �i) The force exerted on segment AB is given by,

FAB = IdL B�

Now, dL = 0.6a x

� FAB = (100 10 ) 0.6 (0.2 0.1 0.2 )3� � �� a a a ax x y z

� FAB = � �12 a 6 a mNy zii) Now the torque on a loop is given by,

T = m B� = IS B�Now area of triangle shaped loop placed in x-y plane can be written in vector form as,

S =12

[0.6 (0.4 )] 0.3a a a ax x y z� � �

Hence T = 100 10 (0.3 ) (0.2 0.1 0.2 )3� � � �� a a a az x y z

� T = 3 a 6 a mN. mx y�

Example 7.5.6

Solution : Consider a circular loop in z = 0 plane

as shown in the Fig. 1.

Current is in a " as shown in the Fig. 1. The given

magnetic field is uniform given by,

B = B02

a ax z�

�

�

� T

The magnetic dipole moment of a planar circularloop is given by,

m = (I S) a n



O

I

ry

z

x

B

Fig. 1

where S is the area of the circular loop.

Note that the loop is laying in z = 0 plane. Thus the direction of unit normal a n must be

decided by the right hand thumb rule. Let the fingures point in the direction of current (in

a " direction), then the right thumb gives the direction of a n which is clearly a z .

� m = I ( ) ( )� �r r I2 2a az z� ... (1)

The total torque is given by,

T = m B� = ( ) ( )� r IB2 0a a az x z� �

2=� r B I2

0

2[ ( )]a a az x z� �

=� r B I

0 0 1

1 0 1

20

2

a a ax y z

=� r B I2

0

2[ ( )]� �a y =

� r B I20

2

��

�

��

a y N.m

Example 7.5.7

Solution : Consider a rectangular

loop in the xy plane with sides b1and b2 as shown in the Fig. 2.

Sides 1 and 3 are of length b1 and

are parallel to y-axis with sides 2

and 4 are of length b2 and are

parallel to x-axis. The origin is at

the centre of the loop.

The vector force on side 1 is given

by,

F1 = I d 1L B�

= [ ]I (b ) (B B B )1 y x x y y za a a a z� � �

= I b [ B B ]1 x z z x� �a a

= I b [B B ]1 z x x za a� ...(1)

The vector force on the side 2 is given by,

F2 = I dL B2 �

= [ ]I ( b ) (B B B )2 x y z� � ��a a a ax x y z

= � �I b [B B ]2 y ya az y

= � � �I b [ B B ]2 z ya ay z ...(2)

The vector force on the side 3 is given by,

F 3 = I d 3L B�

= [ ]I ( b ) (B B B )1 x y z� � ��a a a ay x y z



I

I

I

I

4

3

2

1

Ob2

b2

b1

b1

z

y

x

B

Fig. 2

= � � �I b [ B B ]1 x za az x

= I b [ B B ]1 z x� �a ax z ...(3)

Finally the vector force on the side 4 is given by,

F4 = I d 4L B�

= [ ]I (b ) (B B B )2 x y za a a ax x y z� � �

= I b B B2 y z[ ]a az y�

= I b [ B B ]2 z y� �a ay z ...(4)

Hence total force on the loop of sides b1 and b2 is given by,

F = F F F F1 2 3 4� � �

= I b [B B ] I b [ B + B1 z x 2 z ya a ax z y� � � ]a z

� � � �I b [ B + B ] I b [ B1 z x 2 za ax z a ay z+ B ]y

= I (b B b B b B b1 z 1 x 1 z 1a a ax z x� � �� B )x a z

� � � �(b B b B b B b2 z 2 y 2 z 2a a ay z y �B )y a z

= 0Total torque on the rectangular loop can be obtained by choosing origin of the torque at

the centre of the loop. Hence total torque is given by.

T = T T T T1 2 3 4� � �

But T = R F a a1 1 x x� ��

�

�

� �

b

2I b B2

1 z[ ( B )]x a z

= � �Ib b

2B1 2

x

�

�

� a y ...(5)

T 2 = R F a a2 2 y y� ��

�

�

� � �

b

2I b B1

2 z[ ( + B )]y a z

= �

�

�

�I

b b

2B )1 2

y( a x

= Ib b

2B )1 2

y

�

�

� �( a x ...(6)

T3 = R F a a3 3 x x� ��

�

�

� �

b

2[I b B2

1 z( + B )]x a z

= � ��

�

�

� �I

b b

2B1 2

x a y

= � �Ib b

2B1 2

x

�

�

� a y ...(7)



T4 = R F a ay4 41

2 zb

2I b B� ��

�

�

� �[ ( y za+ B )]y

= � ��

�

�

�I

b b

2B1 2

y a x

= � �Ib b

2B1 2

y

�

�

� � a x ...(8)

Hence adding equations (5), (6), (7), and (8), the total torque is given by,

T = T T T T1 2 3 4� � �

= � �Ib b

2B I

b b

2B1 2

x1 2

y

�

�

� �

�

�

� �a y � �a x

� �Ib b

2B I

b b

21 2

x1 2�

�

�

�

� �

�

�

a y � �� By a x

= Ib b

22 B 2 B ]1 2

y x

�

�

� � �[ a ax y = � �I b b B a B a1 2 y x x y� �

Alternatively we can find total torque by using expression,

T = � �I dS B�Now loop is in x-y plane with dimensions of sides as b1 and b2 .

Then dS = � �b b1 2 a z . Let I be the current. Then total torque is given by,

T = I b b ) B B B )1 2 x y z( (a a a az x y z� � �

= I b b B B )1 2 x y y x(� �a a

= I b b B a B a1 2 y x x y( )� �

Example 7.5.8

Solution : Consider a rectangular loop as

shown in the Fig. 3.

I = 5 A

B = 2 2 4a a ax y z� � Wb/m2

Force on side OA is given by,

FOA = � � �I dB L

= � � ��

�

� �5 (2 2 4 ) (dx )

x 0

x 1

a a a ax y z x

= � � ��5 [ 2 dx 4 dx ]

0

1

a az y = 10 dx 20 dx

0

1

0

1

� ��a az y

= 10 20 N 20 10 Na a a az y y z� � �



x

y

z

O (0, 0, 0) C (0, 2, 0)

A

B

(1, 0, 0)

I

B(1, 2, 0)

Fig. 3.

Force on side AB is given by,

FAB = � � �I dB L = � � �� 5 (2 2 (dy )

y= 0

y= 2

a a a ax y z y4 )

= � ��5 2 dy 4 dy

0

2

a az x = � �� 20 dy 10 dy

0

2

0

2

a ax z = � �40 20 Na ax z

Force on side BC is given by,

FBC = � � �I dB L = � � ��

�

� �5 (2 2 4 ) (dx )

x 1

x 0

a a a ax y z x

= � � ��5 2 dx 4 dx

1

0

a az y = � �� 10 dx 20 dx

1

0

1

0

a az y

= � � �� 10 20 N 20 10 Na a a az y y z

Force on side CO is given by,

FCO = � � �I dB L = � � ��

�

� �5 (2 2 4 ) (dy )

y 2

y 0

a a a ax y z y

= � ��5 2 dy 4 dy

2

0

a az x = � �� 1 dy dy

2

0

2

0

0 20a az x

= � � �20 40 N 40 + 20 Na a a az x x zHence total force on a rectangular loop is given by,

F = F F F FOA AB BC CA� � �

= [20 10 ] [ 40 20 ] [ 20 10 ] [a a a a a ay z x z y z� � � � � � � � 40 20 ]a ax z�

= 0 N

Total torque on a loop is given by,

T = I(d )S B�The loop is placed in in x-y plane i.e. z = 0 plane with dimensions of the rectangular as

1 � 2.

Hence dS = (1) (2) 2a az z�

� T = 5(2 ) (2 2 4 )a a a az x y z� � �

� T = 20 20 0a ay x� �

� T = � � � � � � �20 a 20 a N m 20 ( a a ) N mx y x y



Example 7.5.9

Solution : For N turns loop, the maximum value of magnetic torque is given by,

Tmax = N B I S, where S = Area of a square loop

Now for a square loop, each side is 15 cm i.e. 15 � 10– 2 m

�Tmax = 200 � 1 � 5 (15 � 10– 2 � 15 � 10– 2) = 22.5 Nm

Example 7.7.5

Solution : Let f (x, y) = 2x – 5y. Thus f (x, y) increases from region 2 to region 1 as

region 2 is defined by 2x – 5y < 0 while region 1 by 2x – 5y > 0. Now if we calculate

gradient of f (x, y) then it represents a vector with a magnitude and direction. The

direction of that vector is in the direction of increasing value of f(x,y).

#f =$$

�$$

�$$

fx

fy

fz

a a ax y z

But f = 2x – 5y

� #f = 2 5a ax y�

Now the unit vector normal to the plane is given by,

a n21 =##

��

��

�f

f

2 5

4 25

2 5

29

a a a ax y x y

� a n21 = 0.3714 a x � 0.9284 a y

a) The magnitude of the normal component of B1 is given by,

BN1 = � � � �[ ]B a a1 n21 n12� �But B1 = � � �%1 H H1 1� r1 = ( ) ( )4 10 3 307� � �� a x = 113.0973 � �10 6 a x T

� B1 = 113.0973 � T

b) BN1 = ( )B a a1 21 21� �= � �[ ]113.0973 10 0.3714 0.92846� �� a a ax x y � �0.3714 0.9284a ax y�

= 4.2 10 0.3714 0.92845� �� ( )a ax y = 15.59 a x � 38.99 a y � T

� BN1 = ( ) (15.59 10 38.99 10 )6 2 6 2� � �� = 41.99 � T

c) From the symmetry,

B1 = B Btan1 N1�

� Btan1 = B B1 N1�

= � �[ ]113.0973 15.59 38.99a a ax x y� � � �10 6

= 97.5073 a x � 38.99 a y � T



� Htan1 =� �B a atan1 x y

� � �% r1

97.5073 38.99�

� �

� � �

�

�

10

4 10 3

6

7

� Htan1 = 25.86 a x � 10.34 a y

� Htan1 = � � � �25.86 10.342 2� = 27.8505 A/m

d) According to the boundary conditions,

BN2 = BN1 = 15.59 a x � 38.99 a y � T

Btan2 =��!

&Btan1

� Btan2 =43

10 6[ ]97.5073 38.99a ax y� � �

� Btan2 = 130 a x � 51.98 a y � T

� B2 = B Btan2 N2�

B2 = [( ) ( )]130 10 6a a a ax y x y� � � � �51.98 15.59 38.99

� B2 = 145.59 a x + 12.99 a y � T

� H2 =� �B a a2 x y

� � �% r2

145.59 12.99�

� �

� � �

�

�

10

4 10 4

6

7

� H2 = 28.96 a x + 2.5842 a y

� H2 = � � � �28.96 2.58422 2� = 29.075 A/m

Example 7.7.6

Solution : Given : B2 = 5 8a ax z� mWb m2 , K =1

0�a y mA m.

Now the normal component of B2 is along a z so that the normal component of B1 is also

along a z being continuous at boundary.

By definition,

B1n = B2n = 8a z i.e. Bz = 8

Now for a current at boundary,

� �H H a1 2 n12– � = K

�B B

a1 2z� �1 2

–

��

�

�� =

1� %

a y

�B B

a1 2z� � � �0 1 0 2

–

��

�

�� =

1� %

a y



�B B

a1 2z� �1 2

–

��

�

�� = a y

�� B B Bx y za a a a a

ax y z z z

z

� � ��

�

��

�

�

��

6

5 8

4– = a y

� � �BB

Bxy

z

654

06

84

– – –�

� � � �

�

� �

�

��

�

��a a a ax y z z = a y

� � �BBx

y654

–�

� � �–a ay x = a y

Equation components, By = 0B6

–54

x = –1

�B6x = –1

54

� =14

� Bx =64

= 1.5

Hence B1 = B B Bx y za a ax y z� �

� B1 = 1.5 a a ax y z� �0 8 = � �1.5 a 8 a mWb mx z2�

But H1 =B1

� 1= � �1

80 1� �

1.5a ax z� = � �16

80�

1.5 a ax z�

� H1 = � �10.25 a 1.333 a mA m

0x z��

Example 7.7.7

Solution : z-axis is normal to the boundary. The normal component is given by,

BN1 = � �B1 � �a aN12 N12 � (1)

Here below z = 0, there exists medium 2 while above z = 0, medium 1 exists. The field

vector travels from medium 1 to 2.

� a N12 = –a z

� BN1 = � � � �� 2a – 3a 2a –a –ax y z z z� � � = [–2]� �–a z

= 2 a z mT � Because B1 is expressed in militesla

The tangential component of B1 is given by,

Btan1 = � � � �B B 2 a 3 a 2 a 2 a 2 a – 3 a1 N1 x y z z x y– – –� � � mT.

According to boundary conditions,

BN2 = B 2 aN1 z� mT



Now we can write

� �H Htan1 tan2– = a KN12 � � (2)

But Htan1 =Btan1� �0 1r

=Btan1� 1

=� �

� �2 a 3 ax y� �

� �

10 3–

4 4 10–7�

� Htan1 = 198.94 � �2 a – 3 ax y A/m

Putting value of Htan1 in equation (2), we get,

� �� 198.94 2a – 3a – Hx y tan2 = � �–a z � � �60 a x

� � �397.88 a – 596.82 a Hx y tan2– = –60 a y

� Htan2 = 397.88 a – 536.82 ax y A/m

But Btan2 = � 2 Htan2 = � �7 4 10 397.88 536.82–7� � �� a – ax y

� Btan2 = 3.4999 10 – 4.7222 10–3 –3� �a ax y

� Btan2 = 3.5 4.7222a ax y� mT

Thus B2 = � � � �� B B a – a atan2 N2 x y z� � �3.5 4.7222 2 mT

� B2 = 3.5 a – 4.7222 a 2 ax y z� mT

Example 7.8.8

Solution : For a given solenoid in air,

� = � �0 � � �4 10 7 Wb/A.m

N = 200

d = 6 cm = 6 10 2� � m hence r =d2� � �3 10 2 m

l = 60 cm = 60 10 2� � m

The inductance of a solenoid is given by,

L =� N A2

l=� �0 N ( r2 2 )

l=

� �4 10 200 3 10

60 10

7 2 2 2

2

� � � � � �

�

� �

�

� �( )

= 2.3687 � �10 4 H = 0.2368 mH

Example 7.8.9

Solution : For a toroidal ring of iron,

� r = 800

A = 3 cm 3 10 m2 4 2� � �

R = 10 cm = 10 10 2� � m



N = 500

The inductance of a iron toroid is given by

L =� ��

��

�N A2 R

N A

2 R

20 r

2

�

� L =� � � �

� �4 10 800 500 3 10

2 10 10

7 2 4

2

� � � �

� � �

� �

�

�

�

( )

� L = 0.12 H = 120 mH

Example 7.8.10

Solution : For a co-axial cable, the inductance per unit length is given by,

Ld

=�! �

lnba�

� � =

� �! �% r b

aln�

� � =

4 10 12

3 10

1 10

7 3

3

� � ��

�

�

��

�

��

� �

�

��

ln

= 0.2197 �H/m

Example 7.8.11

Solution : d = 10 m

� r = 80,

a = 1 mm = 1 10 3� � m

b = 4 mm = 4 10 3� � m

The inductance of a co-axial cable is given by,

L =�! �

d ba

ln�

� � =� ��

�% r d b

a2ln �

� � =

4 10 80 102

4 10

1 10

7 3

3

� � � � �

�

��

�

��

� �

�

��

ln

= 2.218 � �10 4 H = 221.8 � �10 6 H

� L = 221.8 �H

Example 7.8.12

Solution : Given :

d = Distance of separation = 2 m

D = Diameter of each wire = 5 cm 5 10 m 0.05 m2� � ��

� a = Radius of each wire =D2

0.025 m�

l = Length of line = 15 m

� r = 1, for air

The total inductance for two parallel wire transmission line is given by,

L =��

� ��

lln

lln

14

d aa

14

d0 r��

�

� �

�

��

��

� aa

�

� �

�

��

��



But d = 2 m >> a = 0.025 m. Hence we get

L =��0 1

4da

lln� �

� �

�

��

��… [� d a d� ' and � r 1� ]

Substituting values,

L =4 10 15 1

42

0.025

7� � ��

�

� �

�

��

��

��

ln = 27.7922 H�

Example 7.8.13

Solution : Given :N1 = N2 = Number of turns = 800

r = Radius of cross section = 5 cm = 5 10 2� �

R = Mean radius of core = 5 cm = 5 10 2� �

� r = Relative permeability = 900

The mutual inductance between closely coupled two toroidal windings is given by,

M =� N N A1 2

l

But A = Area of cross-section = �r 2

and l = Length = Circumference = 2 R�

� M =� � �

�0 r 1 2

2N N ( r )

2 R

Substituting all values, we get,

M =4 10 900 800 800 [ (5 10 )7 2� � � � � � � �� 2

2

]

2 5 10� � ��

��18.0955 H

Example 7.10.3

Solution : The energy density in free space in a magnetic field is given by,

w m =12

H12

H20 r

2� � ��

But for free space � r 1�

� w m =12

H12

4 10 (1000)02 7 2� �� 0.6283 J m3

Example 7.10.4

Solution : Given : N = Number of turns = 500

l = Length of solenoid = 50 cm = 50 10 12� � m = 0.5 m

r = Radius = 10 cm 10 10 2� � m = 0.1 m

� r = Relative pemeability of steel = 3000

I = Current = 10 A



The magnetic field inside solenoid is given by,

H =NI

lA/m

Thus the magnetic flux density is given by

B = � H =( )� �0 r NI

lWb/m2

Hence the total flux " is obtained by multiplying B by the area of cross-section A.

� " = B A� =( )� �0 r NI

Al

� =( )

( )� �

�0 2r NIr W

lb

i) Inductance of the system is given by,

L =NI"

=NI

� ��0 2r NI

rl

( )�

��

��=� � �0

2 2r N r( )

l

� L =4 10 3000 5007 2 2� � � � �� ( ) [ ( ) ]0.1

0.5= 59.2176 H

ii) Energy stored in the system is given by,

WH =12

LI 2 =12

(59.2176) (10)2 = 2.96 kJ

iii) Mean flux density inside solenoid is given by,

B =( )� �0 r NI

l=

4 10 3000 500 107� � � � ��0.5

= 37.69 Wb/m2

��




Example 8.2.8

Solution : To find current in the loop, let us first calculate induced e.m.f.

A circular loop is in z = 0 plane. B is in z-direction which

is perpendicular to the loop. So B is perpendicular to the

circular loop.

Hence total flux is given by

� =

S

d� �B S

With cylindrical co-ordinate system,

dS = (r dr d )� a z

� � = � �� 0.2 sin 10 t (r dr d )

0

2

r 0

0.13

�

�� a az z

� � = � �� 0.2 sin 10 tr2

32

0

0.1

� 02

��

�

��

= � �� 0.2 sin 10 t(0.1)

23

2

2

��

�

��

� � = 6.283 sin 10 t3� �10 3 = 6.283 sin 10 t3 mWb

Now induced e.m.f. is given by,

� e = �ddt�

= � � � �ddt

6.283 sin 10 t310 3

= � � � ��6.283 cos 10 t310 103 3 = � 6.283 cos 10 t3 V

Hence the current in the conductor is given by

i =Induced e. m. f.

Resistance=�6.283 cos 10 t3

5= � 1.2567 cos 10 t3 A


8Maxwell's Equations

and Time Varying Fields

Oy

z

x

0.1 m

B

Fig. 8.1

(8 - 1)

Example 8.2.9

Solution : The inner conductor which is at r1 = 0.03 m rotates at 500 r.p.m. Thus inner

conductor rotates with50060

revolutions per second. As in one second, the distance covered

is (2 r) meter, for the inner conductor the distance covered is50060

��

��(2 r) meters.

Then the linear velocity for inner conductor is given by

v1 = � �� 50060

2��

�� 0.03 a = 1.5707 a � m/s

Similarly for outer conductor, linear velocity is given by

v2 = � �� 50060

2��

�� 0.05 a = 2.6179 a � m/s

Here B is not varying with time, it is constant in a rdirection. Thus under such condition, the induced e.m.f. is

given by,

e = � �E Ld where E = v B�

For inner conductor,

E1 = v B1 1� = � � 1.5707 0.25a a r� � = – 0.3926 a z … a a ar z� � � �

Both the conductors are vertical. Let us assume that length of each conductor be 0.5 m.

Thus, dL1 = dza z

� e1 = � �� E L a a1 1 z zd 0.3926 dz

z = 0

0.5 m

( ) ( ) = – 0.3925 � z 00.5 = – 0.1963 V

For outer conductor,

E2 = v B2 2� = � � 2 0.8.6179 a a r� � = – 2.09432 a z … a a ar z� � � �

� e2 = � �E L1 2d

= � �z = 0

0.5 m

2 dz� � �.09432 a az z = – 2.09432 � z 00.5 = – 1.04716 V

Hence current in the loop is given by,

i =e e

R

0.1963 ( 1.04716)

0.21 2�

��

=0.85086

0.2� 4.2543 A

Example 8.2.10

Solution :

a) e = � ��B

St

d = 4 10 10 103 6 6

0

0 06

0

0 08

( ) ( ) sin..

�

�� t dx dy

xy

Electromagnetic Theory and Transmission Lines 8 - 2 Maxwell's Equations & Time Varying Fields


�

z

R

i

0.5 m

Fig. 8.2

= 4 10 0 08 0 06 103 6� � � �. . sin t = 19.2 sin 106t V

b) e = ( )v�� B Ld = ( ) ( )v a a ay z x�� B dx

x l

0

= �v B l

� �� ( ) ( ) ( . )20 4 10 0 063 = – 4.8 mV

c) e = � � � � �� B

S B Lt

d d( )v

= 4 10 10 10 20 4 103 6 6

00

0 06

� � � � � � ��

�� ( ) sin( ) [

.

t y dy dxy

x

a y� � � �� 3

0 06

0

10cos( ) ] ( )

.

t y dxa az x

= � 240 10 80 1 10 0 06 106

0

3 6cos( ) ( )( . ) cos( )t y t yy

� � � � ��

= 240 10 240 10 4 8 10 106 6 3 6cos( ) cos . cos( )t y t t y� � � � �

= 240 10 240 106 6cos( ) cost y t� � V

Example 8.2.11

Solution : The conductor is placed along x-axis

with one end at origin as shown in the Fig. 8.3.

Given : B = 0.04 a y T

v = Velocity = 2.5 sin 1000 t a z m/sec

By definition, the motional electric field intensity is

given by,

Em = v B�

� Em = (2.5 sin 1000 t a z ) � (0.04 a y )

� Em = 0.1 sin 1000 t ( a )x� V/m … ( a a a )z y x� � � �Now the induced e.m.f. in the conductor is given by,

e = � �� E d ( ) dm L v B L

� e =

x 0

x 0.2

[0.1 sin 1000 t ( )] (dx )

�

�

� � �a ax x = – 0.1 sin 1000 t dx

x 0

0.2

��

� e = � 0.1 sin 1000 t [x]00.2 = – 0.02 sin 1000 t V



x

y

z

0.2 m

v

B

Fig. 8.3

Example 8.2.12

Solution : Here filamentary conductor is fixed and it is

placed in z = 0 plane. It encloses area of 0.65 m2 .

� dS = dS a zInduced e.m.f. according to Faraday's is law is given by,

e = ��

��S

td

BS

= � ��

��

�

��

�

�

��

��

�

��

St

t dS0 05 102

3. cosa a

ay z

z

=� ��

��

�S

tdS

0 05 10 10

2

3 3. sin

= + 35.355 sin 10 3 t

S

dS�

��

�

��

... a ay z� � 0

a az z� �1

But

S

dS� is given as 0.65 m2 .

� e = 35.355 sin 10 3 t (0.65) = 22.98 sin 103 t V

Example 8.3.10

Solution : The ratio of amplitudes of the two current densities is given as 1, so we can

write,

J

J

C

D=

�!

= 1 i.e. � = !

! !"

�r

� � =� ��

2 10

8.854 10 810.2788 10

4

126�

��

�

�rad/sec

But � = 2 f

� f =�#

��0.2788 106

2= 44.372 kHz

Hence, the frequency at which the ratio of amplitudes of conduction and displacementcurrent density is unity, is 44.372 kHz.

Example 8.3.11

Solution : The current through parallel plate capacitor is given by,

iC =!Ad

��

��

dVdt



y

z

x

dS

B

Fig. 8.4

For the given parallel plate capacitor,

! = 100 !0, A = a = 0.01 m2, d = 0.05 mm = 0.05 10 m3� �

V = 100 sin 200 t volts.

Substituting values in the expression for current, we get,

� iC =(100 )(0.01)

(0.05 10 )

ddt

03

!

� �[ sin ]100 200 t

� iC =100 8.854 10 0.01

0.05 10

12

3

� � �

��

�

�100 200 200 cos t

� iC = 11.1262� �10 2003 cos t A …(a)

But for a parallel plate capacitor, iC = iD, thus the displacement current density is given

by,

JD =Displacement current

Area=

i

AD =

i

AC =

11.1262

0.01

� �10 2003 cos t

� J D = 1.1126 cos 200 t A/m2

Example 8.3.12

Solution : Given : iC = 1 A

f = 1 MHz = 1 106� Hz, ! = !0 , = 5.8 107� S/m

Now the current flowing through copper wire is nothing but the conduction current. The

conduction current density is given by,

J C =i

AEC �

� E =i

A1

5.8 10 A

1.7241 10A

C7

8

�

� ��

� �V/m

The displacement current density is given by considering magnitude as,

J D =i

AED � �! ... �

tj j ( )J

DD ED �

��

� �

��

�

�� !

� iD = �! !E A (2 f) ( ) (E A)0�

� iD = 2 1 10 8.854 101.7241 10

A6 12

8� � � � � �

��

A�

� iD = 9.5913 10 A13� �� 0.9591 pA

Example 8.3.13

Solution : Given : = 10 3� S/m

!r = 2.5



E = 5.8 10 sin (9 10 ) t6 9� �� V/m

The conduction current density is given by,

J C = E 10 [5.8 10 sin (9 10 t)]3 6 9� � ��

= 5.8 10 sin (9 10 ) t A / m9 9 2� ��

The displacement current density is given by,

J D =��

��

��

Dt

( E)

t t( E)0 r

!! !

=��

� � � � �� t

[8.854 10 2.5 5.8 10 sin (9 112 6 0 ) t]9

= 8.854 10 2.5 5.8 10 [9 10 co12 6 9� � � � �� s (9 10 ) t]9�

= 1.1554 10 cos (9 10 t) A / m6 9 2� ��

Example 8.3.14

Solution : i) For distilled water :

The ratio of amplitudes of two current densities is unity. Hence we can write,

|J |

|J |C

D=

�!

= 1 i.e. � = !

! !

�0 r

� f�e

=

! ! 25 10

(2 ) (8.854 10 ) (600 r

4

12�

�

�

�

� )= 149.795 kHz

ii) For sea water :

The ratio of amplitudes of two current densities is unity.

Hence we can write,

|J |

|J |C

D=

�!

= 1 i.e. � = !

! !

�0 r

� f�2

=

! ! 23

2 8.854 10 10 r 12�

� � � ��= 53.9265 GHz

Example 8.3.15

Solution : Given : a = Inner radius = 5 mm = 5 10 3� � m

b = Outer radius = 6 mm = 6 10 3� � m

l = Length = 500 mm = 500 10 3� �� 0.5 m

!r = Dielectric constant = 6.7

V = Applied voltage = 250 sin 377 t volts.



Assuming inner conductor at 0 potential and

assuming source free region, the Laplace’s

equation in cylindrical co-ordinate system is given

by,

$ 2 V = 0

Since V is the function of r only, we can write,

1r r

rVr

��

��

��

�� = 0

Since1r% 0; we can write,

��

��r

rVr

��

�� = 0

Integrating both the sides, we get,

rVr

��

= 0 �� A = A

where A is constant of integration

��Vr

=AR

Integrating again, we get,

V =Ar

B�� = A ln [r] + B …(1)

where B is constant of integration.

Using boundary conditions in equation (1) as given below,

At r = a , V = 0

Hence equation (1) becomes,

0 = A ln [a] + B …(2)

Using another boundary condition, at r = b, V = 250 sin 377 t. Hence equation (1) becomes,

250 sin 377 t = A ln [b] + B …(3)

Subtracting equation (2) from equation (3), we get,

250 sin 377 t = A ln [b] – A ln [a] = A lnba ��

� A =250 377sin t

nlba

��

��

=250 377sin t

nl65��

��

…(4)

From equation (2),



ab

0.5 m

Fig. 8.5

B = –A ln [a] = –250 377

5 10 3sin( )

t

n

n

l

l65��

��

� � � …(5)

Hence, V =250 377sin

( )t

n

n r

l

l65��

��

–250 377

5 10 3sin( )

t

n

n

l

l65��

��

� � �

Now, E = �$ V = ��

V

ra r

= ��

��

&

'((

)((

*

+((

,((

��r

t

n

n r250 377

65

sin( )

l

l a r -B is not function of r.Hence neglected

=�

��

��

��

��

250 377sin t

nl65

1r

V ma r

Hence D = ! E =�

��

��

�250 377sin t

nr

l65

!a r C/m2

� D =( ) ( sin )! !0 250 377 1r t

nr

�

��

��

�l

65

a r

Put ! !0 r = 8.854 10 6.712� �� , we get,

D = � � �81.3422 10sin 377 t

r9 a r

Now displacement current density is given by,

J D =��Dt

=�� t

� �&')

*+,

�81.3422 10sin 377 t

r9 a r

J D =� � ��81.3422 10 377

r

9cos 377 t a r =

� � �30.666 10r

6cos 377 t a r A/m2

Hence the displacement current is given by,

iD =

S

d� �J SD =

�

��

�

� ��

�

��

�

�

��

0

2

0

1

377

z

630.666 10

rt r d dzcos (a r a r )

= � � � � �30.666 10 cos 377 t6

= 0

2

z= 0�

�d dz1

= � � �30.666 10 cos 377 t (2 ) (1)6

= � � �192.68 10 cos 377 t A6

Total displacement current = Displacement current per unit length � Length



� iD = [(� � � ��192.68 10 ) cos 377 t ] [500 10 ]6 -3 = � � �96.34 10 cos 377 tA6

= – 96.34 cos 377 t .ANegative sign indicates that the outer conductor is at higher potential than that at

inner conductor. Thus current flows from the outer conductor to the inner conductor. In

dielectric medium, the displacement current is equal to the conduction current i.e. iD = iC.

Example 8.5.11

Solution : According to Maxwell's equation,

$�H = JDt

��

In a free space i.e. air, conduction current density i.e. J is zero.

Hence we can write,

$�H =�� t

D... (1)

L.H.S. of equation (1) can be obtained as follows,

$�H =

a a ax y z��

��

��

� �

x y z

0 0 0.2 cos [210 (3 10 t x)]8

=��

� � ��

��

��y{0.2 cos [210 (3 10 t x)]}

z08 a ax y�

��

� � ��

��

��x

{0.2 cos [210 (3 10 t x)]}z

(0)8

x(0)

y(0)�

��

��

��

��a z

Now in the first term, t and x are constant with respect to y. Hence the derivative is

zero. And the value of last term in equation is also zero.

� $� H =��

� �x

0.2 cos [210 (3 10 t x)]8 a y

= 0.2 { ( 210) sin [210 (3 10 t x)]}8� � � � a y

= (0.2) (210) sin [210 (3 10 t x)] a8y� �

� $�H = 42 sin [210 (3 10 t x)]8� � a y ... (2)

By the definition, the displacement current density is given by,

J D =��

� $ �t

DH ... in free space

Hence from equation (2), we can write,

J D = 42 sin [210 (3 10 t x)] a A / m8y

2� �



Example 8.5.12

Solution : As = 0, we can write

$� H =��Dt

� (1)

Now H = � �2cos 10 t x A m10 – / a z

But B = .H = . .0 r H = � �� 3 10 2 cos 10 t x–5 10� – / a z = � �6 10 cos 10 t x a T–5 10z� – /

L.H.S. of equation (1) is given as,

$� H =

� �

a a ax y z

� � � � � �

/

x y z

cos 10 t x100 0 2 –

= � � � ��

/��

��

/��y

cos t xz x

cos t xz

2 10 0 2 10 010 10– – – – –

��

�� a x � ��

��

� �a ay z0 0

= � �– –2 1010��

/x

cos t x a y = � � � �� – – – –2 1010/ /sin t x a y

= � �– –2 1010/ /sin t x a y … (2)

Equating equation (2) with (1) we get,

��Dt

= � �–2 sin t x/ /1010 – a y

Separating variables,

�D = � �� – –2 1010/ / �sin t x ta y

Integrating both sides with respect to corresponding variables, we get,

D =� �

–2– cos 10 t – x

10

10

10/

/

�

��

�

�

��a y = � �2

10cos 10 t – x a C m

1010

y2/

/

Now, E =D!

=� �� D

a y1.2 10

2

10 1.2 10cos t x

–10 10 –10� �

//1010 –

� E = � �1.6667 cos 10 t – x a V m10y/ /

Example 8.5.13

Solution : For procedure refer example 8.5.12. For above example answers are as follows.

B = � �2 10 cos 10 t – x a T–4 10z� /

D = � � � �1 10 cos 10 – x a C m–9 10ty

2� / /



E = � �8 3333 cos 10 t – x a V m10y� / /

Example 8.5.14

Solution : As 0� �0 0and v , the medium in which D and B are present is nothing

but free space. So the Maxwell's equation obtained from Gauss's law is given by,

$ �D = 0v � 0 ... For free space

��

��

��

D

x

D

y

D

zx y z = 0

��

��

� ��x

(10x)y

( 4y)z

(kz) = 0

� 10 – 4 + k = 0

� k = � 6 .C/m3

Example 8.5.15

Solution : H = Hm cos /z cos �t a y

Now B = . .o r H = . o H = � �. / �o mH cos z cos t a y T

In a charge free region, J � 0

Using Maxwell’s equation,

$� H =��Dt

i.e. � �� $ � H cos tm / �z a y =��Dt

� (1)

Now H has no component in a x and a z , directions. Hence H Hx z� � 0

Thus $�H = – –�

�

�

�

H

z

H

xy y

��

�

��

��

�

��a ax z

Again H is varying with z only and not with x.

��

�

H

xy

= 0

� $�H = –�

�

H

zy

a x = � – cos cos��

/ �z

H z tm a x = � –H t zm cos – sin� / / a x

= � �� / � /H cos t sin zm a x � (2)

Hence equation (1) can be written as,

$�H = � �/ � /H cos t sin zm a x =��Dt

Separating variables,



� �D = � �/ � / �H cos t sin z tm a x

Integrating both the sides with respect to corresponding variables,

D =/�

/ �H

sin z sin t amx

��

��

C/m 2

But

D = ! ! ! !E E E� �o r o � !r = 1 for free space

� E =D H

sin z sin t am

ox!

/� !

/ �o� ��

��

V/m

Example 8.5.16

Solution : a) For time varying fields, we can write Maxwell's equation as,

$�E = ��

Bt

We can write,

$�E =

a a ax y z��

��

��x y z

E E Ex y z

=

�

a a ax y z��

��

��

�x y z

kx 100 t 00

= � � ��

� ��

�z

kx 100 tx

kx 100 ta ax z

Again E is varying with respect to x and not with z.

� $�E = � ��

� � ��x

kx 100 tt

aB

z

� k a z = � �� t

. H = � ��

�.t

x 20 t a z = �20 . a z ... B H� .

Comparing,

k = �20. � �#" (0.5) = – 5 V/m 2

b) Consider Maxwell's equation derived from Gauss's law for electric fields,

$ �D = 0v

��

��

��

D

x

D

y

D

zx y z = 0v = 0 ... Given

From given expressions of D,

Dx = 5x, Dy = – 2y , Dz = kx

Putting values of D , Dx y and Dz , we get,



� � � � � ��

��

� ��x

5xy

2yz

kx = 0

� 5 2 k� � = 0

� k = – 3 .C/m 3

Note that in part (a), k is unknown in the expression of E which is expressed in V/m. In

the expression k is multiplied with x which is expressed in metres (m). Hence accordingly

k is expressed in V/m2 . While in part (b), k is the part of expression of D which is

expressed in . C/m2 . k is multiplied by z which is expressed in m, in expression of D.

Hence k is expressed in .C/m2 .

Example 8.5.17

Solution : From Maxwell's equation, $� E = ��

Bt

�

a a ax y z

� � � � � �x y z(3x kt)0 125 e 0�

= ��

Bt

� � � �&')

*+,

��

��z

125 ex

125 e(3x kt) (3x kt)a x&')

*+,

a z = ��

Bt

The first term will be zero as the function is independent of z, we get

�� x

125 e(3x kt)� ��

��

a z = ��

Bt

375 e(3x kt)� a z = ��

Bt

Separating variables and integrating, we get,

B = � � �375k

e(3x kt) a z

But B = . 0 H … for free space

� H =B. 0

=375

0k.e(3x kt)� a z

From Maxwell's equation based on Ampere's circuit law,

$�H =��Dt

=��

!t

( )0 E … for free space



�

a a ax y z

� � � � � �

.

x y z

0

(3x kt)0375

e0k

�

= !��0 t

E

� �&')

*+,

��

�

��

�� .x k

375e

0

(3x kt) a y = !��0 � t

E

� � �375(3)e

0

(3x kt)k.

a y = !��0 t

E

�� t

E = � �1125

0 0k. !e(3x kt) a y

Separating variables and integration, we get,

E =�

��

��

�

��1125

0 0k. !e

1

k(3x kt) a y

� E =1125

20 0k . !

e(3x kt)� a y V/m

Comparing with original expression,

11252

0 0k . != 125

� k2 =1125

125 0 0( ) ( ). !

� k = 8.99 108�

� H = 1.25 10 e a6 (3x kt)z� � � A/m

Example 8.5.18

Solution : Given : = 0, . r =1.

E = 800 [sin( . )]10 0 016 t z a y� V/m

Comparing with standard field expression,

E = Em [sin( )]� /t z� a y V/m

We get, � = 106 rad/sec, / = 0.01 rad/m, Em = 800.

According to description, given material is with zero conductivity, so we can write,

$� E = ��Bt

� $ ��[ sin( . )]800 10 0 016 t z a y = ��Bt



a a ax y z

��

��

��

"

x y z

sin t z0 800 10 0 016( . )�

= ��Bt

� Simplifying we get,

� � � ��

��z

t zx

t z[ sin( . ) ] [ sin( . ) ]800 10 0 01 800 10 0 016 6a x a z = ��Bt

As function is independent of x, second term is zero.

� � ��z

t z[ sin( . )]800 10 0 016 a x = ��Bt

� [ cos( . )( . )]� � �800 10 0 01 0 016 t z a x = ��Bt

� 8 10 0 016(cos . )t z� a x = ��Bt

Separating variables and integrating with respect to corresponding variables, we get,

8 10 0 016cos( . )t z dt�� a x = �� dB

�8

1010 0 01

66sin( . )t z� a x = �B

� B =�

�8

1010 0 01

66sin( . )t z a x

Now by Maxwell's equation based on Ampere circuital law,

$�H =��Dt

...(� � 0)

But 1 = . . .H H� 0 r = . 0 H ...(. r �1, given)

and D = ! E = ! !0 r E

� $�

��

�

�� B

. 0= � �! !

��0 r tE

� $� �

��8 10

10 0 016

0

6.

sin( . )t z a x = � �! !��0 r tE



a a ax y z

��

��

��

.

x y z

sin t z0 08 10

10 0 016

0

6� ��

�( . )

= � �! !��0 r tE

Simplifying L.H.S., we get,

�� .

�� .y

sin t zx

� ��

��

�

��

�� 8 10

10 0 018 106

0

66

0( . ) a x sin t z( . )10 0 016 �

��

�

��

a y = � �! !��0 r tE

As function is independent of y, the value of first term will be zero on L.H.S. Hence we

get,

��

�

��

�

��

�� .x

sin t z8 10

10 0 016

0

6( . ) a y = � �! !��0 r tE

��

� ��8 10

10 0 01 0 016

0

6.

cos( . )( . )t z a y = � �! !��0 r tE

� � �636 62 10 0 016. cos( . )t z a y = � �! !��0 r tE

Separating variables and integrating with respect to corresponding variables, we get,

� ��636 62

10 0 010

6.cos( . )

! !rt z dt a y = � E�

� E =�

�

�

��

�

��

636 62

8 854 10

10 0 01

1012

6

6

.

.

sin( . )

!r

t za y

� E =� �

�71 9 10

10 0 016

6.sin( . )

!rt z a y V/m

Comparing with given expression, !r = 8.987

��




Example 9.3.6

Solution : From the given expression of E, it is clear that � i.e. phase constant is

associated with x. From the basics of wave propagation, � factor is associated with the

term representing direction of propagation.

Thus, direction of wave propagation is x-direction.

The wave is travelling in free space. In free space,

� = � � ��

From expression of E, term associated with t gives value of � . Hence �� .

� = � �� 10 4 108 7 � �� 8.854 10 12 = 0.3335 rad/m

In free space, the velocity of propagation equals to speed of light in free space.

� =� �

vf

cf

� � 3 10

10 2

8

8 / �= 18.85 m

To calculate the time taken to travel distance � � �, consider expression for velocity.

Velocity =Distance

Time

Time =DistanceVelocity

in free space.

Distance =��

18.852

= 9.425 m

Velocity = c = 3 108 m/s

Hence time taken to travel � � � distance is given by

t =� � �

c9.425

3 108�

= 31.4 nsec


9 EM Wave Characteristics - I

(9 - 1)

Example 9.3.7

Solution : i) For air as a medium, the velocity of proagation is

v = c = 3 × 108 m/s

Then the wavelength is given by,

� =cf

=3 10

10 10

8

6

= 30 m

Hence phase constant � is given by,

� =2��

=230 �

= 0.2094 rad/m

ii) For air, the intrinsic impedance is given by

� = � 0 = 120 � = 377 �The electric field E and the magnetic field H are in phase quadrature. As E is in

x-direction, H must be in y-direction so that the wave travels in z-direction.

H =E�0

=6

377cos (� t – �z) a y A/m

iii) The average Poynting vector is given by

Pavg =12

Re [ ]E H �

We can represent E in phasor form as,

E = 6 ej ( t z)� �� a x

Similarly we can represent H in phasor form as

H =6

377ej t z( )� �� a y

The complex conjugate of H can be written as

H� =6

377e j t z� �( )� � a y

Hence average Poynting vector is given by

Pavg = � �12

66

377e ej t z j t z( ) ( )� � � ��

��

��

!"#

a ax y

=12

36377

$%&

'() (ax ay) = 0.0477 az watt/m2

Example 9.3.8

Solution : E = ( )200 30* + e–j250Z a x

By comparison,

� = 250 =2��

� =2250�

= 0.0251 m

Electromagnetic Theory and Transmission Lines 9 - 2 EM Wave Characteristics - I


f =c�

=3 100 0251

8 .

= 11 9522 109. Hz

But � = 2�f = 2 11 9522 109 � . = 75.0978 GHzIntrinsic impedance of free space is 377 � or 120 �.

H =200 30 90

120

* +, +-

�e–250Z a y = 0.5305 120 e a–250Z

y* + A/m

The wave propagates in positive z-direction.

Example 9.3.9

Solution : E = � �800 cos 10 t y8 �� a z V/m

i) For uniform plane wave in free space,

� =�c

10

3 10

8

8�

= 0.3333 rad/m

ii) � =2 2

0.3333��

�� = 18.85 m

iii) The magnetic field intensity in the free space is given by,

H =E

0�

For free space, � �0 120 377� � �. Since power flow is in y-direction and E is in z-direction,

the direction of H will be + x-direction.

H =� �800 cos 10 t y

377

8 ��a x A/m = 2.122 cos � �10 t y8 �� a x A/m

iv) H at P (0.1, 1.5, 0.4) at t = 8 ns is given by

H = 2.122 cos � � � �� 10 8 10 0.3333 1.58 9 �� a x = 2.122 cos (0.3) a x

H = 2.1219 a x A/m

Example 9.4.8

Solution : E = 10 cos (2 10 t – x)(7� � ,a ay z ) V/m

Given : � = 50 � 0 , � = 2 �0 , . = 0

As . = 0, the medium is lossless medium. For lossless medium, the phase constant � is

given by,

i) � = � �� = ( )2 107� 50 4 10 2 8 854 107 12 � – –. = 2.0958 rad/m

ii) For lossless medium, the intrinsic impedance is given by,

� =��

=50

20

0

��

= 25 0

0

��



But, �0 =��

0

0= 120� = 377

� = 5(377) = 1885 �Now the wave is travelling in +x direction. But E has two components in +y direction and

z direction. So to satisfy the condition of wave travelling in +x direction, H should have

components in +z and –y direction.

Now,

H =E�

=10 2 10

1885

7cos( – )( )� � t x –a az y �a a a

a a a

y z x

z y x

�

�

�

��

�

��and –

Putting value of B

H =10

1885cos(2 10 t – 2.0958x)7� ( )a az y– A/m

H = 5.305 10 3– cos(2 10 t – 2.0958x)7� (a – a )z y A/m

or H = 5.305 cos(2 10 t – 2.0958x)7� (a – a )z y mA/m

Example 9.4.9

Solution : � �H a z� 30 sin 2 10 t – 5x mA m8�

Comparing given expression of H with standard expression, we can write,

Hm = Peak amplitude = 30 10–3 = 0.03 A/m, � � �� 2 10 m sec , 58

Also the wave travels in positive x direction.

For lossless dielectric medium,

� = � �� 0 r 0 r

5 = � � � �� 2 10 4 10 1 8 854 108 7 12� � � – –. r

Simplifying, �r = 5.6915The wavelength is given by,

� =2 2

5��

�� 1.2566 m

The velocity of the wave is given by,

v =��

��

�

2 105

81.2566 10 m sec8

For lossless dielectric medium, the intrinsic impedance is given by,

� =��

� ��

��

�0 r

0 r

–7

–12

4 10 1

8.854 10 5.6915157.9144 � �/158

Now the magnitude of the corresponding electric field is given by,

Em = Hm � �� = 0.03 (158) = 4.74 V/m



As the wave propagates in +ve x direction and H is +ve z-direction the direction of Eshould be a y (so that a a ay z x � ). Hence the corresponding electric field is given by,

E = � �4.74 sin 2 10 t – 5x a V m8y�

Example 9.4.10

Solution : For a lossless medium, . = 0. Then

i) Attenuation constant,

0 = 0ii) Phase constant,

� = � �� 0 0r r

= � � � �� 2 9375 10 4 10 16 7 � �� 8.854 10 2.5612

= 314.37 rad/miii) Wavelength (in polystyrene) is given by,

� =2 2��

�1

�314 37

= 0.01998 m

iv) The velocity of propagation is given by,

v = f � �� 9375 (0.01998) = 1.873 108 m/s

v) The intrinsic impedance is given by,

� =��

� ��

�

��

�

�0

74 10 1r

r 128.854 10 2.56

� = 235.45 �vi) The propagation constant 2 is given by,

2 = 0 , �j = 0 + j 314.37 m 1�

2 = j 314.37 m 1�

vii) The amplitude of electric field intensity is given as 20 V/m. Ex = 20 V/m

But � �EH

x

ywhere Hy is amplitude of magnetic field intensity in y-direction.

� =EH

20H

x

y y� = 235.45

The amplitude of magnetic field intensity is given by,

Hy =E 20

235.45x

�� = 0.08494 A/m

Example 9.4.11

Solution : Given : E = � �5 0 sin 10 t – 4x V m8- a z

Comparing with standard expression of time varying EE = � �E sin t xm � �– a z



We get, Em = 5.0, � = 10 rad sec8 , � = 4 rad m

i) The frequency of propagation is,

� = 2� f

f =��2

=102

8

�= 15.9155 10 Hz6 = 15.9155 MHz

ii) The phase constant is given by,

� = 4 rad miii) The wavelength is given by,

� =cf

=3 10

15.9155 10

8

6

= 18 845 m.

Example 9.5.7

Solution : The propagation constant in lossy dielectric is given by,

2 = � �0 , � � �� ., ��j j j

2 = � �� j 2 f j 2 fr r� � � ., � � �0 0

2 = � � � �j 2 2.56 j 2 8.854 10 , � � �� 10 10 4 10 10 10 109 7 4 9 � �� 12 2.3

2 = � �� j 78.9568 10 2.56 10 j 1.27953 4 ,�

2 = � �� 78.9568 10 90 1.2795 89.983 * + * +

2 = 101.025 10 179.983 * +

2 = 317.84 * +� ,89.99 0.0554 j 317.84

Thus, 0 = 0.0554 Np/m

� = 317.84 rad/m

Example 9.5.8

Solution : The loss tangent for a dielectric is given by,

tan 3 =.��

But tan 3 = 0.005, � = 2.4 �� ... given

0.005 =� ��

.

� � �9 109 2.4 8.854 10 12

. = 6 10 3 � S/m = 6 m S/m

For given dielectric.��j

<< 1, thus it is considered to be practical dielectric.



For practical dielectric,

0 =.�

��

.�

� ��

�� r

r

0 =6 10

24 103 7

� �

�

�

2.4 8.854 10 12= 0.729 Np/m

The propagation constant is given by,

� = � �� 2 f r r

� = � � � �� 2 9 10 4 109 7 � �� 2.4 8.854 10 12

� = 292 rad/mThe wavelength is given by,

� =2 2��

��4�

� = 0.0215 m

Example 9.5.9

Solution : For lossy dielectric medium, propagation constant is given by,

2 = � �j j�� ., ��

2 = � �� j 2 15.9 j 2 15.9 8.854 10 12 , � �� 10 4 10 1 60 106 7 6 � � 50

2 = � �� j 125.5413 60 j 0.0442,

2 = � � � �125.5413 90 60 0.04* + * +

2 = 86.78 45.02* +

2 = 0 , �j = 61.3413 + j 61.3841Comparing real and imaginary terms, we get,

0 = 61.3413 Np / m and

� = 61.3841 rad/m

The velocity of propagation is given by,

v =� ��

�

��

2 10615.9

61.3841= 1.6275 106 m/s

The intrinsic impedance is given by,

� =j

j

��., ��

=� ��

j 2 15.9

j 2 15.9 8.854 10 12

,

�

�

� �

�

10 4 10 1

60 10

6 7

6 � � 50

� =j 125.5413)

60 j 0.0442

(

( ),

� =125.5413 90

60 0.04* +

* +



� = 1.4465 * +44.98 �

Example 9.5.10

Solution : For lossy medium,

2 = j ( j )�� . ��, and

� =j

j

��. ��,

j �� = j(10 3 4 10 )9 –7 � � = j(11.8435 10 3 ) = 11.8435 10 3* 90º

. ��, j = 5 10 2– + j(10 2 8 854 109 12 � . – ) = 5 10 0 055632 ,– .j

= 0.0748* 48.05º

Now the propagation constant is given by,

2 = j ( j )�� . ��, = ( . º )( . . º )11 8435 10 90 0 0748 48 053 * *

= 29.7639* 69.03º

2 = 0 �, j = 10.6518 + j 27.7925

a) 0 = 10.6518 Np/m

b) � = 27.7925 rad/m

c) � =11 8435 10 90

0 0748 48 05

3. º. . º

**

= 397.914 * 41.95º �

d) The velocity of propagation is given by,

v =��

=10

27 7925

9 �.

= 1.1303 108 m/sec

Hence electric field intensity can be given as,

E = 20 e x, 0 cos 10 t x3

9 � ��

, ,$%&

'()a z V/m �(1)

e) E at x = –1.5 m, t = 3 nsec :

Substituting values of x, t, 0, � in equation (1), we get

E = 20 e –1.5, - ,( . )(– . ) –cos ( )( ) ( . )(10 6518 1 5 9 910 3 10 27 7925� ) ,��

��

�3

a z

�(Calculate cos value in radian mode)

= 20 0114 10 0 98026( . )( . )– a z = 2.2348 10 a–6z = 2.2348 a z �V/m

f) H = E� =

20397 914

e x0

.cos 10

341 959 �

��t , ,�

��

x – . a y A/m



At x = –1 m, t = 2 nsec :

H =20

397 914

10 6518 1e( . )(– )

.cos ( )( ) ( . )(– ) –

.–10 2 103

27 7925 141 95

1809 9�

� � , ,

$%&

'()

�

��

��a y

= (0.0502) (23.658 10 6– ) (– 0.6989)a y = – 0.83 a y �A/m

g) For 360º phase shift (i.e. for 2 � radian) the distance travelled by wave is �. Then for 20ºphase shift, the distance d travelled by the wave is given by,

d =20360 �

=20360

2��

$%&

'() =

40360(27.7925)

�= 0.01256 m = 12.56 mm

h) Now amplitude reduces to 12 V/m. Hence we can write,

12 = 20 - e –x0 ( ) because amplitude is reducing

1220

= e(10.6518)(–x)

Taking ln on both the sides,

(10.6518)(–x) = – 0.5108

x = + 47.956 10 –3 m = 47.956 mm

Example 9.5.11

Solution : � r = 1, �r = 48, . � 20 S m, f = 16 GHz = 16 109 Hz

The propagation constant for the lossy dielectric medium,

2 = 0 �, j = j ( j ) j 2 f ( j 2 f )0 r 0 r�� . �� . � � �, � ,

2 = j 2 16 10 4 10 1(20 j 2 16 10 8.854 109 7 9 , � �� 12 48)

2 = j 126.33 10 (20 j 42.7248)3 ,

2 = � � � �126.33 10 90 47.1742 64.93 * + * +

2 = 5.9595 10 154.96 *

2 = 2441.2087 77.45* +

2 = 530.4539 j 2382.8 m 1, �

Example 9.6.9

Solution : The velocity of propagation is given by,

v = f �

f =v��

�

�

2.5 10

0.25 10

5

31 109 Hz = 1 GHz



But the velocity of propagation is also given by,

v =��

��

�2 f

� =2 2 1 109� �f

2.5 105v�

= 25.1327 10 3 rad/m

For good conductor, the phase constant is given by,

� = � �� . � � � .f f r� 0

But for a non-magnetic material, � r �1

� = � �� . �1 10 4 109 7 = 25.1327 10 3

. = 1.6 105 S/m

Example 9.6.10

Solution : If value of factor.��

is very high compared with 1, then the medium is said to

be good conductor. Let us calculate value of above mentioned factor.

.��

=� ��

.�� 0

210

2 60r 128.854 10 15�

�

�= 0.1997 105

The value of ratio is very high compared with 1; thus wet and marshy soil can be

considered as good conductor at 60 Hz.

i) Propagation constant is given by,

2 = j�� .

2 = �� . * +45

2 = � �� 2 60 4 10 1 10 457 2 * +� ��

2 = 2.1766 10 453 * +�

2 = 0 , �j = 1.539 �10 3 + j 1.539 �10 3

Comparing real and imaginary terms,

0 = 1.539 �10 3 Np / m and

� = 1.539 �10 3 rad/m

ii) Skin depth is given by,

5 =10

� �

1

1.539 10 3= 649.75 m

iii) Intrinsic impedance � is given by,

� =1 1.5 .5

, j



� =� � � �

1

10 649.75j

1

10 649.752 2� � ,

� = � �0.1539 j 0.1539, � * +0.21764 45 �

Example 9.6.11

Solution : For conducting medium, . = 58 MS/m. So using expressions of �6 2 and v for

good conductor. The intrinsic impedance is given by

� =j��.

=� �� j f r2 0� � �

.=� �� 2 100 10 4 10 1

58 1090

6 7

6

* +

��

= 3.6896 * +�10 453 �

The propagation constant is given by

2 = j�� . = � �� 2 � � � . 4��f r * + = � �� 2 100 10 4 10 1 58 10 906 7 6 * +��

= 2.1399 * +10 455 Perm

2 = 0 , �j = 1.5131 103 + j 1.5131 105 Perm

Comparing real and imaginary terms, we get,

0 = 1.5131 105 Np / m , and � = 1.5131 105 rad/m

The velocity of propagation is given by

v =��

��

��

�

2 2 100 106f

1.5131 105= 4.152 103 m/s

Example 9.6.12

Solution : For copper, . = 5.8 107 S/m

Then at f = 1 GHz,

.��

=� ��

5.8 10

8.854 10

7

12

�2 1 109�= 1.0425 109 >> 1

Thus at 1 GHz, copper is a very good conductor.

For good conductor,

0 = �� .f

0 = �� 1 10 49 10 5.8 107 7

0 = � = 4.785 �105 m 1

Thus the attenuation constant 0 is 4.785 105 Np/m. The phase constant is 4.785 105

rad/m.

At depth of 0.1 mm, the amplitude of E is given by

7Ex = E ex0z��



Where, Ex0 = Amplitude of E at surface.

z = Depth = 0.1 mm = 1 10 4 � m

7E

Ex

x0= e ez 4.785 105 10 4� � � �� 1.6558 10 21

The phase of E at z = 10 4� m is given by,

Phase = � z = � �4.785 10 105 4 , � = 47.85 rad

Example 9.6.13

Solution : Copper is a good conductor. The depth of penetration for a good conductor is

given by,

5 =� �

1 1

� � . � � � .�f f r

�

For copper, � r = 1 and . = 5.8 107�/m

5 =1

1 10 4 10 16 7� � �

5 = 66.085 � m

Example 9.6.14

Solution : For aluminium with very high conductivity, the propagation constant is given

by,

2 = j��. = j (2 f) ( )0 r� � � .

= j (2 2 10 ) (4 10 1) (40 10 )6 7 6 �� = j 6.3165 108

= 6.3165 10 908 * + = 25.1326 10 45 m3 1 * + �

= 17.771 10 j 17.771 10 m3 3 1 , �

But 2 = 0 �, � , j 17.771 10 j 17.771 103 3

By comparing real and imaginary terms, we get,

0 = 17.771 10 Np m3

� = 17.771 10 rad m3

Hence skin depth for aluminium is given by,

5 =1 1

17.771 10 30��

� 56.2714 m

Similarly the velocity of propagation is given by,

v =��

��

�2 f 2 2 10

17.771 10

6

3� �

� 707.12 m s



Example 9.7.7

Solution : The field vectors can be represented as,

E = 150 sin ( t z)� �� a x V/m

Now H is mutually perpendicular to E and HE

mm

0��

we can write,

H =150�

� �0

( t z)sin � a y A/m

Converting both the sinusoidal functions to cosinusoidal functions,

E = 150 cos t z2

� ��

� �$%&

'() a x and

H =150�

� ��

0t z

2cos � �$

%&

'() a y

Writing in phasor form,

E = 150 ej z( )� �� a x and

H =150�

� � � �

0

j ze ( )� � a y

Thus the complex conjugate of H is given by,

H*

=150

0

2�

��

ej z,$%&

'()

a y

Hence average power density is given by,

Pavg = � �12

Re E H*

= � �12

150150

( ) ( ) ( )�

� � � �

0

j z t + j t ze e$

%&

'

() � � , a ax y

= � �12

150

120

2( )

�a z

= 29.841 a z watt/m2

Now the total power crossing area is given by,

P = � �� P Areaavg

= � � � � � �� 29 84 15 10 30 103 3. � �

= 13.428 m watt

Key Point The flow of power is normal to the area. The area is in z = 0 plane, so the direction

normal to this plane is a z .



Example 9.7.8

Solution : The electric and magnetic fields can be represented in the phasor form as

E = 100 ej

43

x�$

%&

'()

a z and

H =100

120 �

�

ej

43

x$%&

'()

a y

The complex conjugate of the magnetic field H is given by,

H * =100

120 �

�

ej

43

x� $%&

'()

a y

Hence the average power density is given by,

Pavg = � �12

Re *E H

=12

100100

120e e

j43

x j43

x� �

�

$%&

'() � $

%&

'()

�

a az y

�

��

�

�

��

=� � � � � �1

2

100

120

2

�ej 0 �a x

= 13.26 � �� a W / mx2

Now the magnitude of the total power crossing area (path) of 4 m2 of the y-z plane is

given by,

P = (Power density) (Area)

P = (13.26) (4) = 53.05 wattHence the Poynting vector is P = 13.26 � ��a x and the total power crossing area is 53.05

watt.

Example 9.7.9

Solution : E = 1 mV/m

Distance = R = 10 km = 10 m from the station

The magnitude of the magnetic field is given by,

H =E

0�... Power is transmitted in free space

=1 10

120

3 �

�

= 2.65 10 6 � A/m



The magnitude of the power density is given by

P = (E) (H) = � �� 1 10 2.65 103 6 � � � �2.65 10 W / m9 2

The power is transmitted in radial direction at a distance 10 km from the station over aspherical region. The area of the spherical region is given by

S = � �4 r 4 10 12.5663 10 m2 4 2 9 2� ��

Thus the total power radiated over the spherical region is given by,

Power = (Power density) (Area)

= � �� 2.65 10 12.5663 109 8 �

= 3.33 W

Example 9.7.10

Solution : i) For air as a medium, the velocity of propagation is,

v = c = 3 × 108 m/s

Then the wavelength is given by,

� =cf

=3 10

10 10

8

6

= 30 m

Hence phase constant � is given by,

� =2��

=230 �

= 0.2094 rad/m

ii) For air, the intrinsic impedance is given by,

� = �o = 120 � = 377 �The electric field E and the magnetic field H are in phase quadrature. As E is in

x-direction, H must be in y-direction so that the wave travels in z-direction.

H =E�o

=6

377cos (� t – � z) a y A/m

iii) The average Poynting vector is given by,

Pavg =12

Re [ ]E H �

We can represent E in phasor form as,

E = 6 ej ( t z)� �� a x

Similarly we can represent H in phasor form as,

H =6

377ej t z( )� �� a y

The complex conjugate of H can be written as,

H� =6

377e j t z� �( )� � a y

Hence average Poynting vector is given by,

Pavg = � �12

66

377e ej t z j t z( ) ( )� � � ��

��

��

!"#

a ax y



=12

36377

$%&

'() (ax ay)

= 0.0477 az watt/m2

Example 9.7.11

Solution : i) The resistance of a metallic conductor is given by

R = 8l

A

but 8 =1 1

2 107.�

l = Length = 1m

A = Area of circular cross-section = � �r (1 10 )2 2 2� �

Substituting the values of 8, l and A, we get,

R =1

(2 10 ) ( ) (1 10 )1.5915 10

7 2 24

�

��

��

Hence the d.c. power loss in conductor is given by

Ploss = I R (100) (1.5915 10 )2 2 4� � = 1.5915 W

ii) The current density is given by

J =Current

Area

J =100

1 10 2 2�( ) �

J = 318.309 10 A / m3 2

As the current is flowing in z-direction, we can write

J = 318.309 10 A / m3 2 a z

The conduction current density is given by

J = . E

E =J 318.309 10

2 10

3

7.�

a z

E = 0.0159154 V / ma 15.9154 10 a V / mz3

z� �

The magnetic field within a cylindrical conductor of radius 'a' carrying current I is given

by

H =Ir

2 a 2�9a ; r < a



H =100 r

2 (1 10 )2 2�9

�a

H = 153.15 r a k A / m9

The Poynting vector is given by,

P = E H

= (15.9154 10 ) (159.15 r 10 )3 3 � a az 9

= 2533 r ( a ) W / mr2�

iii) In the cylindrical co-ordinate system, the surface area in the a r direction is given by,

d S = r d dz9 a r

where r is radius of a conductor

s

d: -P S =

z 0

1

0

2

[2533 r ( )] [r d dz

� �: : � -

9

�

9a r a r ] = �� : :2533 r d dz2

z 0

1

0

2

9

�

9

= � 2533 (r ) [ ] [z]202 p

019

= � �2533 (1 10 ) [2 ] [1]2 2 �

= – 1.5915 W

��




Example 10.2.7

Solution : For medium 1, i.e. free space,

�1 = 120 � �� For medium 2, i.e. perfect dielectric,

�2 =� �

� �

�

� �

��

��

�

�r

r 128.854 10

4 10 1

3

7

� �2 = 217.507 �

The transmission coefficient is given by,

� =�

� EE

2 217.507

377 217.507t

i�

��

�

2 2

1 2

��

= 0.7317

The reflection coefficient is given by,

� =� �

EE

217.507 377

217.507 377r

i�

��

��

� ��

2 1

2 1= – 0.2683

In terms of magnetic field magnitudes, we can write,HH

t

i=

E

EEE

t

i

t

i

/

/

��

��

2

1

1

2� �

��

��

�HH

t

i=

377217.507

(0.7317) = 1.2682

SimilarlyHH

r

i= �

��

E

EEE

0.2683r

i

r

i

/

/

��

1

10.2683

Example 10.2.8

Solution : The interface is between perfect dielectric (region 1) and free space (region 2).

For region 1,

�1 =�

1

1=

4 10 17

�

�

�

8.8542 10 8.512= 129.22 �


10 EM Wave Characteristics - II

(10 - 1)

For region 2,

�2 = 120 � = 377 �By definitions,

� =EE

t

i�

�2 2

1 2

��

=�

� 2 377

129.22 377�= 1.4894

But Ei = 1.5 mV/m

� Et = � � � � E 1.4894 1.5 10i3� �

= 2.2341 mV/m

� =� �

EE

377 129.22

377 129.22r

i�

��

��

� ��

2 1

2 1= 0.4894

But Ei = 1.5 mV/m

� Er = � � � � E 0.4894 1.5 10i3� � = 0.7341 mV/m

As we know, Ht =Et

�2and H

Hr

r��1

we can write,

Ht =2.2341 10

377

3 �= 5.9259 A/m and

Hr =0.7341 10

129.22

3 �

�= – 5.681 A/m

Example 10.2.9

Solution : As medium 1 is free space,

�� = 120 � = 377 �For medium 2, � � � indicates it is lossless

dielectric. For lossless dielectric, intrinsic

impedance is given by,

�2 =�

� ��

�� r

r

� �2 =4 10 97

�

�

�

8.854 10 412= 565.1 �


� =� 2 2

2

��

��

2 565.1

377 565.1= 1.1996


� =� ��

�

�

2

2

��

��

565.1 377565.1 377

= 0.1996

Electromagnetic Theory and Transmission Lines 10 - 2 EM Wave Characteristics - II


Region 1

= 8.5

= 1

= 0

= –––

�

�

��

r1

r1

1

1

�

1

1

Region 2

Freespace

= 377� �2

Fig. 10.1

Medium 1Free space

Medium 2Material with

= 4, = 9

and = 0

�

�r r

Fig. 10.2

Example 10.2.10

Solution : The interface is between perfect dielectric

(region 1) and free space (region 2).

For region 1,

�1 =�

1

1=

4 10 17

�

�

�

8.8542 10 8.512= 129.22 �

For region 2,

�2 = 120 � = 377 �

By definitions,

� =�

� EE

2 377

129.22 377t

i�

��

�

2 2

1 2

��

= 1.4894

But Ei = 1.5 mV/m

� Et = � � � � E 1.4894 1.5 10i3� � = 2.2341 mV/m

� =� �

EE

377 129.22

377 129.22r

i�

��

��

� ��

2 1

2 1= 0.4894

But Ei = 1.5 mV/m

� Er = � � � � E 0.4894 1.5 10i3� � = 0.7341 mV/m

As we know, Ht =Et

�2and H

Hr

r��1

we can write,

Ht =2.2341 10

377

3 �= 5.9259 A/m and

Hr =0.7341 10

129.22

3 �

�= – 5.681 A/m

Example 10.3.3

Solution : A uniform plane wave in free space is propagating in positive x-direction. It

strikes a large block of copper lies in y-z plane. Let E be in y-direction and H be in

z-direction. Let Ei be the amplitude of the electric field in the incident wave. Consider the

equation for the electric field in the incident wave given by,

E = � E t zi cos � �� a y V/m ... (1)

The interface is between free space and copper which is good conductor as shown in the

Fig. 10.3.5.



Region 1

= 8.5

= 1

= 0

= –––

�

�

��

r1

r1

1

1

�

1

1

Region 2

Freespace

= 377� �2

Fig. 10.3

For medium 1, i.e. free space,

�� = � � ��0 120�For medium 2, (for copper)

�2 =��

� �45 ... As copper is good conductor

=� � 2 300 10 4 10 1

45

6 7

� �

��

5.8 107

= 6.3905 10 453 � �� = � 4.5187 10 j 4.5187 103 3 � � � �


� =2 2

2

��

�EE

t

i

Et =�

� E

2 6.3905 10 45

377 4.5187 10 j 4.5187 10i

3

3 3

� �

� �

�

� �

!

""

#

$

%%

� Et = E0.0127 45

377 j 4.5187 10i 3

� �

�

!""

#

$%%�

� Et =0.0127 45

377 6.89 10E

4 i�

�

!"

#

$%�

� Et = � 3.3687 10 45 E5i � �� V/m ... (2)


� =EE

r

i�

��

� ��

�

�

2

2

� Er =4.5187 10 j 4.5187 10 377

4.5187 10 j 4.5187

3 3

3

� �

�

� �

� 10 377E

3 i� �

!""

#

$%%

� Er =� �

�

!""

#

$%%

�

�

377 j 4.5183 10

377 j 4.5183 10E

3

3 i

� Er =377 180

377 6.89 10E

4 i� �

�

!"

#

$%�

� Er = & '1 180 E Ei i� � � � ... (3)

The amplitude of the magnetic field in the incident wave in medium 1 is given by,

Hi = � E E377

2.6525 10 Ei i 3i��

� � � A/m ... (4)

The amplitude of the magnetic field in the transmitted wave in medium 2 is given by,



Ht =& '�

E 3.3687 10 45 E

6.3905 10 455.2714 10t

5i

3�2�

� �

� ��

�

� � �3iE A/m ... (5)

The amplitude of the magnetic field in the reflected wave in medium 1 is given by,

Hr = � E E377

2.6525 10 Er i 3i�

��

� � ��

A/m ... (6)

As the wave is travelling in free space, the velocity equal to velocity of light in air.

� v = c = 3 108 ��

� � =� �c

�

2 300 10

3 10

6

8= 6.2831 rad/m ... (7)

Using equations (1) to (7), the time domain expressions of incident, reflected and

transmitted waves are as follows.

Incident wave :

Ei = � E t 6.2831 xi6cos 2 300 10 �� a y V/m

Hi = � � 2.6525 10 Ei t 6.2831 x3 �� cos 2 300 106� a z A/m

Transmitted wave :

Et = � � 3.3687 10 E t 6.2831 x 455i � � �� cos 2 300 106� a y V/m

Ht = � � 5.2714 10 E t 6.2831 x 453i � � �� cos 2 300 106� a x A/m

Reflected wave :

Er = � � �E 10 t 6.2831 xi6cos 2 300� a y V/m

Hr = � � � ��2.6525 10 E 10 t 6.2831 x3i

6cos 2 300� a z A/m

Example 10.3.4

Solution : The conductivity of silver is given as � = 61.7 M S/m >> 1, it can be

considered as good conductor. Thus the intrinsic impedance for good conductor is given

by,

�( =��

� �45 =� � �

�� r f2

45� �

=� �

� 4 10 1 2 15 10

45

7 6

� �

��

61.7 106

= 1.3854 10 453 � �� = � 9.7962 10 j 9.7962 104 4 � � � �

As medium 1 is free space, the intrinsic impedance is given by,

�� = � �� 377 0 � � � �377 j 0

Also in free space, Ei = 100 V/m




� =EE

t

i=

2��

(

� (

�E100

t =�

� � 2 1.3854 10 45

377 + j 0 9.7962 10 j 9.7962 10

3

4 4

� �

� �

�

� �

� Et =0.2771 45

377 j 9.7962 10 4

� �

� �

� Et =0.2771 45

377 1.48 10 4

� �

� �= 7.35 10 454 � �� V/m


� =EE

r

i=

� ��

( �

( �

�E100

r =� 9.7962 10 j 9.7962 10 377

9.7962 10 j 9.7962

4 4

4

� �

�

� �

�� 10 3774� �

� Er = 100� �

�

!""

#

$%%

�

�

377 j 9.7962 10

377 j 9.7962 10

4

4

� Er = 100377 180

377 1.48 10100 180

4

��

�

!"

#

$% � � �

�

� Er = – 100 V/mNote that 180� angle represents that the wave propagates in opposite direction to that of

the incident wave. Er is almost equal to Ei indicates that the wave cannot propagate in

good conductor but gets totally reflected.

��




Example 11.7.4

Solution : For a transmission line,

Z0 = Z ZOC SC � � �� 550 500 60 14

= 524 404742

524 404 37. .�� = 418.807 – j 315.594 �

Now tanh � =Z

Z500 14550 60

0.953414SC

OC�

��

� �� 60

2

= 0 9534 23. � � � 0.877 j 0.3725

tanh � =e

e

2

2

1

1

�

�

�

e

e

2

2

1

1

�

�

�

= 0.877 + j 0.3725

e2 1� � = � � � �0.877 j 0.3725 e 2 1�

e2 1� � = 0 877 2 2. e j 0.3725 e 0.877 j 0.3725� �

�e 1 0.877 j 0.37252 � � � = 1 0.877 j 0.3725

� �e 0.123 j 0.37252 � � = 1.877 j 0.3725

e2 � =1.877 j 0.3725

0.123 j 0.37251.9136 11.22

0.

�

��

3922 71.72��

e2 � = 4 8791 82 94. .� �

2 � = �ln 4 8791 82 94. .� �

� =12

4 8791 82 94ln[ . . ]� �

Mathematically, �ln lna b a j b� �


11 Transmission Lines - I

(11 - 1)

� = � �12

ln 4.8791 j 82.94 �

But � = � �j

� �j =12

4 8791ln . ��

��

��j

82.942

� =12

4 8791ln . � 0.7924 Nepers/km

and � =82 94

241 47

..

��

�� /km = 0.7237 rad/km

Note : Another mathematical result can be used to solve this problem.

tanh � = tanh � �� j A j B�

then � =12

1tanh �

��

��

2A

1 A B2 2Nepers

and � =� �

12

1tan �

�

�

��

��

�

��

��

2 B

1 A B2 2

Units of � depends on tan �1 function mode. If tan �1 is calculated in degree mode, will be

in degrees and if tan �1 is calculated in radian mode, � will be in radians.

Example 11.7.5

Solution : The value of Z0 is,

Z0 = Z Z 720 34 570 48SC OC � � ��

= 640.6246 �� 7 �

and tanh (� l) =Z

Z720 34570 48

SC

OC�

� ��

= 1.1239 41 0.8482 j 0.7373� � �

Now tanh (� l) =e

e0.8482 j 0.7373

2

2

1

1

�

�

l

l

�

�

ez l� �1 = � � � �e 0.8482 j 0.73732 1� l

�e 1 0.8482 j 0.73732 � l � � = 1 0.8482 j 0.7373

e2 � l =1.8482 j 0.7373

0.1518 j 0.73731.9898 21.74

��

� 80.7527 78.36

��

e2 � l = 2.6435 100.108 2.6435 1.747� �� rad

2� l = ln [ 2.6435 � 1.747 rad ]

Now ln [a<b] = ln a + j b

Electromagnetic Theory and Transmission Lines 11 - 2 Transmission Lines - I


2� l = ln 2.6435 + j 1.747

� = �1

2 lln2.6435 j 1.747 but l = 10 km

� = 0.0486 + j 0.08735

Example 11.9.9

Solution : L = 0.56 mH/km, C = 0.1 F km/

Leakage conductance and conductor resistance are negligible.

Z0 =ZY

=R j L

G j C

!!

"LC

... as R << j !L and G << j !C

Z0 =0 56 10

01 10

3

6

.

.

�

�

�

�= 74.8331 �

Similarly, � = ZY = (R j L)(G j C) ! ! " j LC2 2! = j L! C

But � = � � j = j! ( . ) ( . )0 56 10 01 103 6� �� = j km! ( . )7 4833 10 6 1� � �

Comparing imaginary terms,

� = ( . )7 4833 10 6� � ! rad/km

Thus the phase velocity is given by,

v =!�

=1

7 4833 10 6. � �= 1.3363×105 km/sec

Example 11.9.10

Solution : !� #$ � #$f � �1 106 rad/sec

R + j ! L = 83 4 2 10 2 29 10 143887366 3. . . � � � � � ��j 89.667$

G + j ! C = 0 + j 2 10 4 85 10 10 0 304736 2 6$� � � � � � �� . . 90

Z0 =R j L

G j C14388.736 89.667

0.30473 90

��

!!

= 217.297 �� 0.166 �

And � = � �� R j L G j C 14388.736 0.30473 89.667 90 � � � � �! !

= 66.2169 89.833 0.1924 j 66.216 j� � � � � �

� = 0.1924 neper/km and � = 66.216 rad/km

% =2 2

66 216

$�

$�

.= 0.0948 km



v =!�

$�

� �2 1 10

66 216

6

.= 94889.23 km/sec.

Example 11.9.11

Solution : It is known that,

Z0 =R j L

G j C

!!

and � = � �� R j L G j C ! !

Z0 � = R j L !

R j L ! = 692 12 0.0363 78 25.1196 66�� = 10.217 j 22.9478

R = 10.217 �/km

!L = 22.9478

L =22 9478

222 9478

2 1000. .$ $f

��

= 3.6522 mH/km ... f = 1 kHz

and�

Z0= G j C !

G j C ! =0.0363 78692 12

0.000052456 90� �

�� = 0 + j 0.000052456

G = 0 mho/km

! C = 0.000052456

C =0 000052456

2 1000.$�

� � �8.3486 10 9 F/km

� = � �j = 0.007547 + j 0.0355

� = 0.0355 rad/km

% =2 2

0 0355

$�

$�

.= 176.957 km

v =!�

$�

�2 1000

0 0355.= 176957.45 km/sec

Example 11.9.12

Solution : Given : f = 1000 Hz

R = 6 km� L = 2.2 mH/km

G = 0.25 10 6� ��/km C = 0.005 10 F km6� �

Series impedance Z R j L� ! = R j (2 f) L $

= 6 j (2 1000) (2.2 10 )3 � � � �$



= 6 + j 13.823 = 15.069 � 66.53º

Shunt admittance Y = G + j ! C = 0.25 10 j (2 1000) (0.005 10 )6 6� � � �� $

= 0.25 10 j 31.4159 106 6� ��

= 3.1416 10 89.545� � ��

i) The characteristic impedance is given by,

Z0 =ZY

15.069 66.53

3.1416 10 89.545�

� �

� � ��

�692.575 11.51��

ii) The propagation constant is given by,

� = ZY (15.069 66.53 ) (3.1416 10 89.54 )5� � � � � ��

= 0.02175 78.04 km 0.0045 j 0.02127 km1� � � � � 1

Example 11.9.13

Solution : For a transmission line of length l = 2 miles,

R = 30 �/mile L = 2.2 mH/mile

G = 20 n�/mile, C = 80 nF/mile and

f = 10 kHz

The characteristic impedance Z0 is given by,

Z0 =R j L

G j C

!!

=30 2 10 10 2 2 10

20 10 2 10 10 80

3 3

9 3

� � � � �

� � � � � �

�

�

j

j

( ) .

( (

$

$ 10 9� )

=30 138 23

20 10 5 0265 109 3

� ��

j

j

.

.=

141 4479 77 75

5 0265 10 89 993

. . º

. . º

�

� ��

= 167.7511 � – 6.12 �

The propagation constant � is given by,

� = (R j L)(G j C) ! !

= ( . ) (30 2 10 10 2 2 10 20 10 2 10 10 803 3 9 3 � � � � � � � � � � �� j j$ $ 10 9� )

= (141.4479 77.75º ) (5.0265 10 89.99º )3� � ��

= 0.8432 � 83.87º / km

Representing � in rectangular form, we get

� = � + j � = 0.09 + j 0.8383

Hence attenuation constant = � = 0.09 nepers/miles

Phase constant � = 0.8383 rad/miles



Example 11.9.14

Solution : The Z0 is given by,

Z0 =R j L

G j C

!!

where !� # $ $f � � �2 1 10 3 rad/sec.

Z0 =� �10.4 j 2 10 0.00367

0.8 10 j 2 10 0.00835 1

3

6 3

� �

� � � ��

$

$� �0 6�

=10.4 j 23.059

0.8 10 j 5.246 10

25.29 656 5

� ��

��

.72

5.246 10 89.1265

�

� � ��

= 4 8208 10 6572 891265. . .� � ��

Z0 = 694.32 �� 11 703. �

Now � = � �� R j L G j C � � �� ! ! 25 29 6572 5 246 10 891265. . . .

= 0 001326 154 846. .� �

= 0 03641 77 423. .� � = 0.007928 + j 0.03553 = � + j �

� = 0.007928 nepers/km and � = 0.03553 radians/km

% =2 2

0 03553

$�

$�

.= 176.841 km

v =!�

$�

� ��

2 1 10

0 03553

3

.1.95 104 km/sec.

Now Z0 =E

IS

S... ES = 1 V

I S =E

Z1 0

694.32 11.703S

0�

� ��

= 1.4402 10 11.7033� � �� A

Example 11.13.2

Solution : For a line short circuited at the receiving end,

E

IS

S= � �Z0 tanh � l

ES = � �I ZS 0 tanh � l ... (1)

And I = � � � �I xE

ZxS

S

0cosh sinh� ��

And x = l = 20 km at the receiving end

I = � ��

� �II Z

ZSS 0

0cosh

tanhsinh�

��l

ll� ... using (1)



= � ��

I IS Scoshsinh

cosh�

�

�l

l

l�

2

=� � � � �

� �I S

cosh sinh

cosh

2 2� �

�

l l

l

�=

� �I S

cosh � l

Now I S = 20 mA, � = 0.1 + j 0.05

I =� � � �

20 10

cosh 0.1 j 0.05 20

20 10

cosh 2 j 1

3 3�

��

�

� �=

20 10cosh 2 cos 1 j sinh 2 sin 1

3�

�

... use radian mode

=20 10

2.0327 j 3.0518920 10

3.667 0.9832 rad

3 3�

��

�

� �

= 0.00545 0.9832 rad�� 5.45 56.33 mAThus received voltage is zero and received current 5.45 mA lags the sent current by 56 33. �.

��




Example 12.2.4

Solution : Given R = 0.006 �/m, L = 2.5 � 10-6 H/m, C = 4.45 pF/ m, f = 10 MHz. At

f = 10 MHz, � �L fL� 2 = 2 10 10 2 5 106 6� � � � � �� . = 15.708 �. Hence �L >> R at 10 MHz.

So according to standard assumption for the dissipationless line, we can neglect R.

i) The characteristic impedance is given by,

Z0 = R0 =LC

=2 5 10

4 45 10

6

12

.

.

�

��

�

�749.53 �

ii) The propagation constant is given by,

� = � j = 0 � j LC�

Hence � = � j = � 0 2 10 10 2 5 10 4 45 106 6 12� � � � � � �� j � . .

� � = � j = 0 + j 0.2095 per m

� Attenuation constant = = 0 and Phase constant = = 0.2095 rad/m

iii) The velocity of propagation is given by,

� =1

LC=

1

2 5 10 4 45 106 12. .� � �� = 2.998 108� m/sec

iv) The wavelength is given by,

� =2�

=2

0 2095�

.= 29.9913 m

Example 12.2.5

Solution : For high frequency line, f = 10 MHz, C = 0.01 �F/km, L = 3 mH/km.

Z0 =LC

=3 10

0 01 10

3

6

�

�

�

�.= 547.72 �

� = � � j = j LC� = j � 3 10 0 01 103 6� � �� .

� � = j km� ( . )5 4772 10 6 1� � �


12 Transmission Lines - II

(12 - 1)

� � = j ( ) ( . )2 10 10 5 4772 106 6� � � � �� ... ( )�� 2 f

� � = j 344.1426 km–1

Example 12.5.8

Solution :

The characteristic impedance of line is given by,

Z0 = RLC

9 10

16 100

6

12� �

�

��

�

�750 � ... at R.F. only

Hence reflection coefficient is given by,

K =Z Z

Z Z

1000 750

1000 750R 0

R 0

��

��

� 0.1428

The standing wave ratio S is given by,

S =1 K

1 K1 0.14281 0.1428

��

��

�| |

| |1.3333

Example 12.5.9

Solution : At radio frequencies, SR

R1200R

0

R R� � for resistive load

K =s 1s 1

0.2��

�

� K =

1200R

1

1200R

10.2R

R

�

��

Simplifying for RR , we get

RR = 800 �

Example 12.5.10

Solution : The reflection coefficient is given by,

K =� � � �

Z Z

Z Z

300 j 400 300

300 j 400 300

j 40R 0

R 0

��

��

� ��

0

600 j 400�

� K =400 90

721.11 33.69

� ��

Electromagnetic Theory and Transmission Lines 12 - 2 Transmission Lines - II


ESER R =1000R �

�S �R

Fig. 12.1

� K = 0.5547 56.31� �The standing wave ratio S is given by

S =1 |K|

1 |K|

��

=1 0.55471 0.5547��

= 3.4913

Example 12.5.11

Solution : Given : Z0 = R0 = 100 � , ZR = 200 �


K =Z Z

Z ZR 0

R 0

��

=200 100

200 100

��

=100300

= 0.3333

The standing wave ratio is given by,

S =1 |K|1 |K|��

=1 0 3333

1 0 3333

��

.

.= 1.99985 � 2

Example 12.5.12

Solution : VSWR =1 |K|1 |K|

1.5��

�

� 1 + |K| = 1.5 – 1.5 |K|

� 2.5 |K| = 1.5 – 1 = 0.5

� |K| = 0.2

Example 12.5.13

Solution : Special considerations of radio frequency line

Refer section 12.1.

Given : Z0 = 70 � ZR = 115 – j 80

The reflection coefficient K is given by

K =Z Z

Z ZR 0

R 0

��

=(115 j 80) 70

(115 j 80) 70

� ��

=45 j 80

185 j 80

��

=91.7877 60.64

201.5564 23.38

o

o

��

��= 0.4553 �� 37.26o

Hence VSWR is given by,

S =1 |K|

1 |K|

��

=1 0.45531 0.4533��

= 2.6717

Maximum line impedance is given by,

Zs(max) = S Z0 = S R0 = (2.6717) (70) = 187.02 �Minimum line impedance is given by,

Zs(min) =Z

S0 =

R

S0 =

70(2.6717)

= 26.2 �



Example 12.9.3

Solution : For a quarter wave transformer, the input impedance is given by,

Zin = ZR

ZS02

R�

The source impedance ZS � 500 �

Load impedance = ZR � 200 �

� 500 =R

20002

� R02 = (500) (200)

� R02 = 100000

� R0 = 316.22 �

The frequency of operation is

f = 200 MHz

Hence the wavelength is given by,

f . � = c

� � =cf

3 10

200 10

8

6�

�

�= 1.5 m

� The length of quarter wave line is given by,

s =��

4

1 5

4

.= 0.375 m

Hence the quarter wave line is as shown below in the Fig. 12.2

Example 12.9.4

Solution : Given : ZR �150 � , R0 �75 � , f = 12 GHz

For a quarter wave transformer, the input impedance is given by,



Z = 500in �

Z =200R �

s = = 0.375 m�

––4

Fig. 12.2

Zin = � ZR

Zor R Z ZS

02

R0 S R� � � �

� Zin =�

Z75

15037.5S

2

� � �

Thus �R0 of the matching section is 37.5 �

The operating frequency is 12 GHz.

� The wavelength can be calculated as,

f . � = c

� � =cf

3 10

12 10

8

9�

�

�= 0.025 m = 25 cm

Hence the length of quarter wave line is given by

s =��

40.025

4= 6.25 cm

Thus the quarter wave transformer is as shown in the Fig. 12.3

Example 12.9.5

Solution : For quarter wave transformer,

Zin = Zs =R

Z

R

Z02

R

02

L�

75 =R

02

100

� R02 = 7500

� R0 = 86.6 �



Load

Z =150R �

s = = 6.25 cm�

––4

R' = 37.50 �

The quarter wave line

Line with

R = 750 �

Fig. 12.3 A quarter wave matching section

Example 12.10.7

Solution : Given : Zo = 50 �, ZR = (75 – j 20) �, f = 100 MHz

� � =cf

=3 10

100 10

8

6

�

�= 3 m


K =Z Z

Z ZR 0

R 0

��

=(75 j 20) 50

(75 j 20) 50

� ��

=25 j 20

125 j 20

��

=32.0156 38.66º

126.5898 9.09

��

= 0.2536 � – 29.57º

� K = |K| � � = 0.2536 � – 29.57º

Hence |K| = 0.2536 and � = – 29.57º = – 0.5161 rad

Location of short circuited stub :

d =��4

[� + � – cos– 1 |K|] =��4

� ��

� � ��

��

� !

"#$

�[cos (0.2536)]1180

=3

4�[– 0.5161 + � – 1.3144] = 0.313 m

Hence the short circuited stub must be located 0.313 m or 31.3 cm from load.

Length of short circuited stub :

L =��2

1tan � �1 |K|

2|K|

2

=3

21

�tan � ��

%

!%

"#%

$%

1 (0.2536)

2(0.2536)

2�

180��

�� = 0.5194 m

Hence the length of the stub must be 0.5194 m or 51.94 cm.

Example 12.10.8

Solution : Given : Zo = 75 �, ZR = (150 + j 225) �, f = 500 MHz

� � =cf

=3 10

500 10

8

6

�

�= 0.6 m


K =Z Z

Z ZR 0

R 0

��

=(150 j 225) 5

150 j 225 5

� ��

7

7=

75 j 225

225 j 225

��

=237.1708 71.56º

318.198 45º��

= 0.7453 � – 26.56º

� K = |K| � � = 0.7453 � – 26.56º

Hence |K| = 0.7453 and � = – 26.56º = – 0.4635 rad

Location of single shorted stub

d =��4

[� + � – cos– 1 |K|] =0 64.�

� � � ��

��

� !

"#$

�0.4635 cos 0.7453]1��

[180

=0.64�

[– 0.4635 + � – 0.7298] = 0.093 m



Hence the stub must be located at a distance 0.093 m or 9.3 cm from load.

Length of single shorted stub :

L =��2

1tan � �1 |K|

2|K|

2

=0.62

1 (0.7453)

2(0.7453)

2

�tan � ��

%

!%

"#%

$%

1 �180

��

�� = 0.0401 m

Thus the length of the single shorted stub must be 0.0401 m or 4.01 cm.

Example 12.10.9

Solution : Given

R0 = 600 � , ZR = 1800 �The reflection coefficient is given by,

K =Z R

Z RR

R

��

��

� � � �0

0

1800 6001800 600

12002400

0.5 0

Hence � = 0, |K| = 0.5

Calculation for the location and length of stub :

Case (1) : s1 =� ��

cos 1

2

|K|

But =2 ��

� s1 =� ��

�

��

�

��

��

��

cos |1

22

K|=

� 0 0 5

4

1��

��

cos .= 0.1666 �

� L =� �&

'

((

)

*

++

�2

1�

tan1 |K|

2 |K|

2

=� �

� �&

'

((

)

*

++

�2

1�

tan1 0.5

2 0.5

2

= 0.1135 �

Case (2) : s1 =� ��

� ��

cos 1

4

|K|=

� 0 0 5

4

1��

��

cos .= 0.3333 �

� L1 =� �

�

&

'

((

)

*

++

�2

1�

tan1 |K|

2 |K|

2

=� �

� �

�

&

'

((

)

*

++

�2

1�

tan1 0.5

2 0.5

2

= � �

��2

1�

tan 0.866 = � � �2

1�

�� tan 0.866 = 0.386 �

Example 12.10.10

Solution : The normalized load admittance is given as,

Y

GR

0= 1.25 + j 0.25

�1 / Z

1 / RR

0= � 1.25 j 0.25�



�R

Z0

R= � 1.25 j 0.25�


K =Z Z

Z Z

Z R

Z R

1R

Z

1R

Z

R 0

R 0

R 0

R 0

0

R

0

R

��

��

��

�

� K =� �

1 1.25 j 0.25

1 1.25 j 0.25

� �

� �=

� ��

0.25 j 0.25

2.25 j 0.25=

0.3535 1356.34

�� 2 2638.

= 0.1561 �� 141.34

= 0.1561 �� 2.466

Calculating value of � cos�1 K ,

� � cos�1 |K| = � cos� �1 0.1561 1.414

Calculation for the length and location of the stub :

Case (1) : s1 =� ��

cos 1

2

|K|=

� ��

�

��

�

��

��

��

cos 1

22

|K|=

� ��

2 466

4

. ��,-.,.�

= � �0 0587.

Length of the stub,

L =� �&

'

((

)

*

++

�2

1�

tan1 |K|

2 |K|

2

=� �

� �&

'

((

)

*

++

�2

1�

tan1 0.3535

2 0.3535

2

= 0.1469 �

Case (2) : s1 =� ��

� ��

cos 1

4

|K|=

� � ��

2.466 3.142 1.4144 �

= 0.1662 �

Length of the stub,

L =� �

�

&

'

((

)

*

++

�2

1�

tan1 |K|

2 |K|

2

=� �

� �

�

&

'

((

)

*

++

�2

1�

tan1 0.3535

2 0.3535

2

= � �

��2

1�

tan 1.3231

= � / 0� �2

1 0132311�

�� tan . = 0.353 �

Example 12.10.11

Solution : Given

f = 150 MHz, R0 = 600 �, ZR = 75 �Calculating � first,

f. � = c



� =cf

3 10

150 10

8

6�

�

�= 2 m

The reflection coefficient is given by

K =Z Z

Z Z

Z R

Z RR 0

R 0

R 0

R 0

��

��

� K =75 60075 600

525675

0.7777��

��

� �

� K = 07777 07777 180. .� � � ��c

The two possible locations of stub are as follows.

Case (1) : s1 =� � � ��

cos 1

2

|K|

But =2 ��

� s1 =� � �

� ��

�cos 1

4

|K|=

� �

� � ��

� �cos 1

42

0.7777= 0.8918 m

The length of the stub is given by

L =� � �&

'

((

)

*

++

�2

1�

tan1 |K|

2 |K|

2

� L =� �

22

1�

tan � �&

'

((

)

*

++

1 0.7777

2 0.7777

2

= 0.1222 m

Case (2) : s1 =� � �

��

�cos 1

4

|K|=

� � ��

��

��cos 1

42

0.7777= 1.108 m

The length of the stub is given by

L =� � �

�

&

'

((

)

*

++

�2

1�

tan1 |K|

2 |K|

2

=� �

� �

�

&

'

((

)

*

++

�2

1�

tan1 0.7777

2 0.7777

2

= � 2

21

�tan � � 0.4041 = � 1

0 384�

�� . = 0.8777 m

Selecting a point located nearest to the load. Hence the stub location nearest to the load is

calculated in case (1).

The stub must be located at a distance 0.8918 m from the load and the length of the stub

required is 0.1222 m.



Example 12.12.7

Solution : [A] Calculate first normalized impedance.Z

ZL

0=

Z

RR

0=

100 200

100

� j= 1 – j 2

Locate point A at the intersection of constant resistance circle of 1 and capacitive reactance

circle of – 2.

Draw a circle passing through point A which represents standing wave ratio.

For given line, � =cf

=3 10

10 10

8

6

�

�= 30 m

Now from load point A, move towards generator in clockwise direction. The distance to be

covered is given by

s1 = 25 m =2530

� = 0.8333 �

Thus locating point B at a distance 0.8333 � from point A.

At point B susceptance is + 1.2 and resistance is 0.44

Thus the impedance of a line at a distance 25 m from load is,

ZS = R0 (0.44 + j 1.2) = 100 (0.44 + j 1.2)

� ZS = (44 + j 120) �To calculate input impedance, travelling 100 m 1 3.3333 � distance from load point A. It is

observed that we reach at same point. Thus the input impedance is again same.

� Zin = (44 + j 120) �[B] To calculate input admittance, let us first calculate normalized admittance.

Y

GR

0=

1Z

1Z

R

0

=Z

Z0

R=

100100 j 200�

=1

1 j 2�

�Y

GR

0=

1+ j 2

5= 0.2 + j0.4

Locate point L as shown on the Smith chart. It is a load point. Travelling 100 m i.e. 3.3333

� distance from point L we reach at input point M.

At point M conductance is 0.28 while susceptance is – 0.75. Hence the admittance at the

input is given by

Yin = G 0 (0.28 – j 0.75) = (0.28 – j 0.75)

Yin = 2.8 10 j 7.5 103 3� � ��



Example 12.12.8

Solution : [A] The successive voltage minima or maxima are separated by distance�2

always.

��2

= 102 cm

� � = 204 cm = 2.04 m



Fig. 12.4

Hence frequency of operation can be calculated as,

f � � = c

� f =c 3 10 m / s

2.04 m

8

��

�= 147.058 MHz

The load impedance in terms of standing wave ratio S and distance of first minima �s is

given by

ZL = R

1 j Ss

S j tan2 s0

��

��

��

�

��

��

��

��

�

��

&

'

((((

)

*

tan2 �

�

++++

Here R0 = 250, S = 2.4, �s = 35 cm, � = 204 cm

� ZL =�

250

2 35204

2 35204

1 j 2.4

2.4 j

��

��

��

��

��

�

tan

tan

�

�

��

&

'

((((

)

*

++++

� ZL =� �

� 250

1 j 2.4 tan 1.078

2.4 j tan 1.078

c

c

�

�

&

'

((

)

*

++

� ZL = 2501 j 4.469

2.4 j 1.8622

��

&

'()

*+

� ZL = 2503 0377 3

4.5795 77.387.8

��

&

'()

*+.

� ZL = 376 88 39 58. .�� ZL = � 290.47 j 240.13� �[B] Let us solve using Smith chart.

S = 2.4

First draw a circle with chart centre (1, 0) as centre of circle and 2.4 as radius. This is

S-circle with S = 2.4.

This circle cuts real axis on right hand side of the centre at a point whose co-ordinates are

(2.4, 0). This circle cuts at point A on left hand side of centre on the real axis. This point A

is the point of voltage minima. Co-ordinates of the point A are (0.42, 0).

The last minimum is 35 cm away from the load.

The total length is 204 cm.

Hence the load is35

20401716

�� . away from last minimum.

So if we travel a distance 0.1716 � from the point A, we will get a point B which is

nothing but the load point. Note that we are moving from a voltage minima to load i.e.

from generator to load, hence move in counterclockwise direction.



The coordinates of B are, (1.15, – 1.0)

Hence normalized load impedance is given by,Z

RL

0= � 1.15 j 1.0�

� ZL = � R 1.15 j 1.00 �

� ZL = � 250 1.15 j 1.0�

� ZL = � 287.5 j 250� �



Fig. 12.5

Example 12.12.9

Solution : 1. The normalized load impedance is given by,

ZR =Z

Z

25 j50

50L

0�

�= 0.5 + j1

Locate point A at intersection of r = 0.5 circle and X = + j 1 circle. This is a load

point.

2. With origin O as center. Draw a circle with radius equal to OA. This is constant S

circle. Extend line OA to point A' at periphery.

3. Constant S circle drawn in step 2, cuts horizontal axis at point P on the right hand

side. Through this point r = 4.2 circle passes, hence

S = VSWR = 4.24. Now with radius OA draw arc DE on the reflection coefficient line located at the

lower side of chart. This arc cuts line at F. At point F1 value of K can be obtained.

And it is

K = 0.63Now the angle correspond to point A' is 83º. This is the angle of reflection coefficient

is given by,

K = K � � = 0.63 � 83º5. Now from point �A , move distance 3.3 � in clockwise direction (load to generator) to

achieve input point. On the periphery we must complete 6 rotations to achieve 3 �distance and then travel 0.3 � distance to reach input point B'. Join point OB' which

intersects constant-S circle at point B. It is the intersects of r = 0.8 circle and

X = – 1.4 circle.

Thus normalized input impedance is

Zs = 0.8 – j 1.4

Hence actual input impedance is given by,

Zs = Z0 (Zs) = 50(0.8 – j 1.4) = (40 – j 70) �Refer Fig. 12.6 on next page

Example 12.13.5

Solution : ZR = 50 – j 100, Z0 = R0 = 50 �

The normalized load admittance is given by,

Y

GR

0=

1Z

1Z

R

0

=Z

Z0

R=

5050 j 100�

=1

1 j 2�=

1 j 2

5

�= 0.2 + j 0.4

Locate point A on the chart indicating the normalized admittance.



Draw constant S circle through point A. It cuts the ra axis at 5.8 which indicates standing

wave ratio before the use of stub is 5.8.



Fig. 12.6



Fig. 12.7

Locate point B at the intersection of S circle and unity conductance circle. Now this point

is nearest to the load, At B the susceptance is + 2.0 which is capacitive in nature. This

point gives the position at which the stub is to be connected.

Extending line through point A and B from the centre of the chart upto the outermost rim

of the chart, we get s = 0.062 � for point A and s = 0.188 � for point B. Travelling from

load to generator in clockwise direction, the distance from the load at which stub is to be

connected is given by,

s1 = [0.188 � – 0.062 �] = 0.126 �As the line susceptance is capacitive, the line must provide inductive susceptance of –

2.0. Locate point C at the intersection of – 2.0 susceptance circle with the outer rim of the

chart. Thus at C, S = 0.322 �. As the stub is short circuited at other end, measuring

distance from extreme right hand point on the real ra axis i.e. from short circuit point

(S = 0.25 �). Thus the length of the stub required is given by,

L = (0.322 � – 0.25 �] = 0.072 �For matching load of (50 – j 100) � with 50 � line, a short circuited stub must be located

at distance 0.126 � from the load. The length of the stub required is 0.072 �. Refer Fig. 12.7.

Example 12.13.6

Solution : Using admittances as short circuited stub is connected in parallel.

The normalized admittance can be obtained as,

Y

GR

0=

1Z

1Z

R

0

�

��

�

��

�

��

�

��

=Z

Z0

R=

75150 225� j

=75

270 41 56 31. .� �= 0.2773 �� 56.31º

�Y

GR

0= 0.1538 – j 0.2307

a) Locate point A at [0.1538 – j 0.2307] on chart as shown. It indicates load point.

b) Draw S.W.R. circle with centre at O(1, 0) and radius equal to OA.

c) The SWR circle cuts unity conductance circle in points B and C which give possible stublocations.

d) Draw lines from O through A, B and C.

e) Consider point C first. Location of stub from load point is S1.

S1 = Distance between radial lines OA and OC going towards generator in clockwise

direction.

= {(0.5 – 0.465) � + 0.194 � } = 0.229 �

f) The line susceptance is inductive. Hence stub should provide capacitive susceptance. Thesuspectance of line at point C is + j 2.4. The stub should provide – j 2.4 curve intersects it.Hence extend line OD. Then the length of the stub is measured from extreme right handpoint on real axis. Thus length of the stub is



L1 = {0.312 � – 0.25 � } = 0.062 �

Example 12.13.7

Solution : Given : Load impedance ZL = 75 – j 20 �

Characteristic impedance Z0 = 50 �

1. Normalized load impedance is given by

ZL =Z

Z

75 j 20

501.5 j 0.4L

0�

��

Refer Fig. 12.9 on next page.

The impedance ZL is located at point A. Exactly opposite to point A, a normalized

load admittance point B is located. Extend OB to point B' on the periphery.



Fig. 12.8

2. Intersection of constant - S circle drawn through point A and g = 1 circle is shown as

point C and E . But print C is nearest to load . Hence neglecting point E.

3. Join OC and extend it to point C' located at periphery. Now the distance between B'

and C' gives the location of a stub from load i.e.

d = 0.148 � – 0.0355 � = 0.1125 � m



Fig. 12.9

4. The susceptance at point C is the susceptance of line at stub connection. It is equal to

j 0.58. It is capacitive susceptance. It is neutralized by inductive susceptance of

– j 0.58.

5. The point corresponding to – j 0.58 is plotted on g = 0 circle and b = – 0.58 at point

D' as shown.

6. Point P represents short circuit point in admittance chart. Hence distance of point D'

from P gives length of stub i.e.

l = 0.415 � – 0.25 � = 0.165 � m

Example 12.13.8

Solution : ZR = 50 – j 100, Z0 = R0 = 50 �

The normalized load admittance is given by,

Y

GR

0=

1Z

1Z

R

0

=Z

Z0

R=

5050 j 100�

=1

1 j 2�=

1 j 2

5

�= 0.2 + j 0.4

Locate point A on the chart indicating the normalized admittance.

Draw constant S circle through point A. It cuts the ra axis at 5.8 which indicates standing

wave ratio before the use of stub is 5.8.

Locate point B at the intersection of S circle and unity conductance circle. Now this point

is nearest to the load, At B the susceptance is + 2.0 which is capacitive in nature. This

point gives the position at which the stub is to be connected.

Extending line through point A and B from the centre of the chart upto the outermost rim

of the chart, we get s = 0.062 � for point A and s = 0.188 � for point B. Travelling from

load to generator in clockwise direction, the distance from the load at which stub is to be

connected is given by,

s1 = [0.188 � – 0.062 �] = 0.126 �As the line susceptance is capacitive, the line must provide inductive susceptance of –

2.0. Locate point C at the intersection of – 2.0 susceptance circle with the outer rim of the

chart. Thus at C, S = 0.322 �. As the stub is short circuited at other end, measuring

distance from extreme right hand point on the real ra axis i.e. from short circuit point

(S = 0.25 �). Thus the length of the stub required is given by,

L = (0.322 � – 0.25 �] = 0.072 � .

For matching load of (50 – j 100) � with 50 � line, a short circuited stub must be locatedat distance 0.126 � from the load. The length of the stub required is 0.072 �.



Example 12.14.2

Solution : The normalized load admittance is given by

YG 0

=

1Z

1Z

R

0

=Z

Z0

R=

5035 j 35�

=1

0.7 j 0.7�



Fig. 12.10

�Y

G 0= 0.7142 – j 0.7142

Locate point O on the chart as shown. Stub1 adds a susceptance and this must changeY

G 0

to a valueY

G1

0which will lie on locus of all points i.e. circle B. Stub1 cannot alter

conductance. Hence following a constant conductance circle from point O locates point 1

where,Y

G1

0= 0.714 – j 0.49

The stub1 must contribute a susceptance of ba = + j 0.224 = j 0.23; at 1, 1 on the line.

The [+ 0.23] constant susceptance circle cuts the rim of the chart at 0.036 � .



Fig. 12.11

The stub1 at 1, 1 is shorted. Hence starting from extreme right hand end (short circuit) and

travelling in clockwise direction.

Length of the stub at 1, 1= L1= (0.5 – 0.25) � + 0.036 � = 0.286 �

The section of the line between points 1,1 and 2,2 changes the admittance at 1, 1 to that of

2, 2 along a constant circle with 0 as centre and 0-1 as radius.

The admittance at 2, 2 on the line without the stub can be read from chart at point 2 on

the circle A givingY

G2

0= 1 + j 0.68

Hence stub-2 should then be adjusted to provide an inductive susceptance of

ba = � 0.68 after which the admittance at 2, 2 will beY

G2

0= 1 + j 0 and the main line to

the left of terminal 2, 2 will be properly terminated.

The [– 0.68] circle cuts the outer rim at 0.404 � . Then the length of the stub2 is given by

L2 = 0.404 � – 0.25 � = 0.154

��



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