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Gauss’ Law Author: Pranjal K. Bharti (IIT Kharagpur) Mb: 7488044834

P a g e | 1 CONCEPT: JB-20, Near Jitendra Cinema, City Centre, Bokaro www.vidyadrishti.org

2013-2015

Target Boards, JEE Main & Advanced (IIT), NEET 2015

Physics Gauss’ Law

Author: Pranjal K. Bharti (B. Tech., IIT Kharagpur) H. O. D. Physics, Concept Bokaro Centre

© 2007 P. K. Bharti

All rights reserved.

www.vidyadrishti.org

Gauss’ Law Author: Pranjal K. Bharti (IIT Kharagpur) Mb: 7488044834

P a g e | 2 CONCEPT: JB-20, Near Jitendra Cinema, City Centre, Bokaro www.vidyadrishti.org

O

E y

x 45o

Prerequisite

• Vectors • Newton’s Laws • Coulomb’s Law • Electric Field Please revise these topics before studying Gauss’s Law

Introduction

• We are going to introduce the concept of Gauss’s Law which can be used for symmetrical situation with utmost ease. Using Gauss’ Law we can we find electric field in just two-three steps for a symmetrical situation.

• Concept of area vector and flux is an important part of Gauss’s Law which we will introduce before introducing Gauss’s Law.

Area Vector

• Let us start with an example of square surface of area A. Area vector A

corresponding to this square is a vector perpendicular to its surface having magnitude equal to its area A. Thus,

A A=

• Again let us see an example of a cube of edge a. Clearly,

surface area of each face is a2. Thus, a cube have six area vectors corresponding to each side as shown in the figure. Clearly,

2Top Bottom Left Right Front BackA A A A A A a= = = = = =

. • One important thing to note here is that each area vector

is taken in the outward normal direction. • These were standard figures, where we can define area

vector in (outward for a closed surface) normal direction. All faces were flat. What if we get an irregular surface?

• For the case of an irregular surface we divide the surface into infinitesimally small areas. Let us consider such a small area to be dA. Then, area vector corresponding to this small surface is d A

in the outward normal direction as shown in the figure.

• We can find out total surface area (in scalar form) of an irregular surface using integration. Total surface area is given by

A d A d A= =∫ ∫

Area Vector (Summary)

Area vector of a flat surface is a vector perpendicular to surface having magnitude equal to its area A.

Area vector is taken in outward normal direction for a

closed surface.

We can find out total surface area (in scalar form) of an irregular surface using integration.

Quick Exercise 1

1. Consider the given cube of edge 3m in given coordinate system. Write area vectors of all six faces in unit vector notation.

2. Suppose a uniform electric field E exists in a region along +ve x-axis. Cross section of a flat surface having area 2√2 m2 at an angle 45o with x-axis is shown in figure. Find area vector in unit vector notation.

Answers: 2 2 1. Face ABEF: 9 , face OCDG: 9 ,m i m i−

2 2Face BCDE: 9 , face OAFG: 9 ,m j m j−

2 2Face DEFG: 9 , face OABC: 9m k m k−

( ) 2 2. 2 2i j m−

O A

B C

D E

F G

z

y

x

Gauss’s Law Author: P. K. Bharti (IIT Kharagpur) Mb: 7488044834

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Electric Flux

• Physical meaning of Flux: Roughly speaking, we can picture flux φ in terms of number of electric field lines passing normally through an area A.

• Increasing the area means that more lines of E

pass through the area, increasing the flux.

• Stronger field means more closely spaced lines of E

and therefore more lines per unit area, so again the flux increases.

• Definition (For Boards): The total number of electric field lines passing normally through an area placed inside an electric field is called electric flux.

• Expression for flux: Electric Flux through a flat surface of area A, when uniform electric field E passes through it in any direction is given by

.E Aφ =

(Electric flux through a flat area in a uniform field) Conditions for this expression:

a) Surface is flat & b) Electric field is uniform

Flux for general case:

• We have already defined flux through a flat surface in a uniform electric field. What happens if the electric field E

isn't uniform but varies from point to point over the area A? Or what if A is part of a curved surface?

• We divide surface area A into many small elements dA, with area vector .d A

• We calculate the electric flux .E d A

through each element and integrate the results to obtain the total flux:

E d Aφ = ⋅∫ (Electric flux for general case)

For a completely closed surface we can use closed integral ∫ in place of ∫ . Circle over integral sign is

used to denote integration over a closed surface. E d Aφ = ⋅∫

(Electric flux through a closed surface)

Example 1

• Find the net flux through a closed cylinder of length l and radius r in a uniform electric field E parallel to its axis.

Solution:

• The closed cylinder is composed of three surfaces: left circular surface, right circular surface and a curved surface. Thus, net flux through the cylinder , φnet

= (flux through right circular surface, φleft)

+ (flux through left circular surface, φRight)

+ (flux through curved surface, φCurved).

i.e., φnet = φLeft + φRight + φCurved …(1)

• Let us calculate flux through each surface.

Flux through right circular surface:

• Area vector RightA

for right surface must be taken in outward normal direction, i.e., towards right.

• As the right surface is flat and field is uniform, we can

use .E Aφ =

• Thus, flux through right circular surface

rightright right. cos 0 ...(2)E A EA E rφ π= = = 2

Flux through left circular surface:

• Area vector leftA

for left surface must be taken in outward

normal direction, i.e., towards left. • Thus, flux through left circular surface

left

2leftleft . cos180 ...(3)E A EA E rφ π= = = −

Flux through curved surface: • As this surface is not flat we can’t use .E Aφ =

• We will have to use, . .E d Aφ =

• Let us take a small element of area d A

on the curved

surface. Clearly, angle between E

and d A

is 900 .

• Thus, flux through this small area,

. cos90 0od E d A EdAφ = = =

. • If we take any such element on the

curved surface, area vector d A

will be perpendicular to

E

. Hence, flux due to all such elements on the curved surface = 0.

E



• Hence, net flux through curved surface,

Curved . 0 ...(4)E d Aφ = =∫

• Using eqns. (1), (2), (3) & (4), we get, φnet = φleft + φRight + φCurved φnet = 0

• Hence, flux through a closed cylinder in a uniform electric field = zero.

DISCUSSION: • We have seen from above example that flux through a

closed cylinder in a uniform electric field is zero. • This is true for all closed surface completely inside a

uniform electric field whether it is cylinder or not. • We know that the net flux through the surface is

proportional to the net number of lines leaving the surface, where the net number means the number leaving the surface minus the number entering the surface.

• Uniform electric field means that electric field lines are parallel and spacing between any two field line is same. Hence, if a closed surface is placed in a region of uniform electric field, the net number of lines leaving the surface is zero, since same number of field lines leave as that which enter.

• Thus, flux through a closed surface in a uniform electric field is zero.

Quick Exercise 2

NCERT Questions

1.15. Consider a uniform electric field 33 10 N/CE i= ×

(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 600 angle with the x-axis? 1.16. What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? More questions for practice

1. A circular plane of radius 10 cm is placed in a uniform electric field of 5× 105 NC-1, making an angle of 600 with the field. Calculate electric flux through the sheet.

2. ( )If E 6 i 3 j 4 k N/C,= + +

calculate the electric flux

through a surface of area 20 m2 in Y-Z plane. 3. In a region of the space, the electric field is in the x-

direction and proportional to 0, . ., .x i e E E xi=

Consider an imaginary cubical volume of edge a, with its edges parallel to the axes of coordinates. Find the flux through this volume.

4. An electric field in a region is given by

1200 i NC x > 0E for−=

and 1200 i NC x < 0E for−= −

A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = + 10 cm and other is at x = – 10 cm.

i. What is the flux through each flat face? ii. What is the flux through the side of the cylinder?

iii. What is the net outward flux through the cylinder?

5. In a region of the space, the electric field is in the x-

direction and proportional to

0, . ., .x i e E E x=

Find the flux through the shaded area. (IIT-JEE 2011)

Solutions:

NCERT Questions:

1.15. (a) 30 Nm2 C-1 (a) 15 Nm2 C-1

1.16. zero

More questions for practice

1. 1.36 × 104 Nm2 C-1 2. 120 Nm2 C-1 3. 3

0E a 4. i. 1.57 Nm2 C-1 through each flat face

ii. zero iii. 3.14 Nm2 C-1

5. 20E a



Gauss’ Law

• In this section we describe a general relationship between the net electric flux through a closed surface (often called a Gaussian surface) and the charge enclosed by the surface.

• Definition: Gauss' law states that the total electric flux

through any closed surface is 0

1ε

times total charge enclosed

by that surface. Thus,

( )enc

0

Gauss' Lawqφε

=

Important points to note about Gauss’s Law: 1. φ = electric flux through closed surface because of electric

field created by all charges (inside and outside the Gaussian surface)

2. qenc

= total charges which are inside the Gaussian surface.

Using definition of flux through a closed surface, we can write Gauss’s Law as

( )enc

0

. Gauss' Lawq

E d Aε

=∫

Alternate definition of Gauss’s Law: The surface integral of

electric field through any closed surface is 0

1ε

times total

charge enclosed by that surface. • Gauss's law is an alternative to Coulomb's law. Gauss’s law is

more fundamental law than Coulomb’s Law. It is very useful in symmetric situations.

Derivation of Gauss’s Law

• Let us consider a point positive charge Q. • Let us consider the total flux through a sphere of radius r,

which encloses this point charge Q. • Now, let us consider a small

surface of area dA and area vector d A

on the surface of the sphere.

• Clearly, angle between E

and d A

is 00. Thus, flux through this small area,

0. cos 0 .d E d A EdA EdAφ = = =

If we take any such element on the curved surface, area vector d A

will be parallel to .E

Hence, net flux due to all such elements on the Gaussian surface .

...(1)

E d A EdA

E dA

φ

φ

= =

⇔ =

∫ ∫∫

• We have taken E outside integral because magnitude of

electric field is same everywhere on the surface of the Gaussian surface. Now,

( ) ( )2 22

0

22

0

0

4 44

surface area ofspherical surface 4 &4

QE dA E r rr

QdA r Er

Q

φ π ππε

ππε

φε

= = =

= = =

⇒ =

∫

∫

• This proves Gauss’s law.

CCoonnssttrruuccttiioonn ooff GGaauussssiiaann SSuurrffaaccee

Point charge: • We should use symmetry to construct a Gaussian surface.

The most appropriate Gaussian surface for a point charge is a sphere. The point charge Q lies at centre of the sphere.

• The most appropriate Gaussian surface for an infinite line charge is a cylinder. The line charge lies on the axis of the cylinder.

• An appropriate Gaussian surface for an infinite plane charge is a cylinder. Axis should be perpendicular to the surface.

AAPPPPLLIICCAATTIIOONN OOFF GGAAUUSSSS’’SS LLAAWW

• Step by step method to find electric field using Gauss’s law:

1. Gauss’s Law is generally used for symmetrical situation. First construct a closed Gaussian surface using symmetrical situation.

2. Draw electric field lines. 3. Take a small area vector d A

perpendicular to surface at one or more points according to situation.

4. Find flux through this closed surface using .E d Aφ = ∫

Most of the cases we will find that E

and d A

are either parallel or anti-parallel for an element of area dA. Thus, we will get .E d A EdA=

when &E d A

are parallel and .E d A EdA= −

when &E d A

are anti-parallel. 6. Find the total charge q

enc inside the Gaussian surface.

Finally use Gauss’s Law enc

0

.q

E d Aφε

= =∫

and get

electric field magnitude.

r

Q d A

E

Q

λ σ

Point charge

Infinite Line charge

Infinite Surface charge



AApppplliiccaattiioonn ooff ggaauussss’’ss llaaww:: FFiieelldd dduuee ttoo aa ppooiinntt cchhaarrggee

• Let us consider a point positive charge Q. Suppose we have to find out electric field at point P located at a distance r from this charge using Gauss’s Law.

• The Gaussian surface is in the form of a sphere. The radius of spherical Gaussian surface should be r, because we are interested to find out electric field at a distance r from the charge Q.

• We know that electric field due to a point positive charge in the radial outward direction. Also the magnitude of electric field at any point on the Gaussian surface is same because every point is at same distance r from the point charge Q.

• Now, let us consider a small surface of area with area

vector d A

on the surface of the sphere. Clearly, angle between E

and d A

is 00 at each point on the Gaussian surface. Thus, flux through this small area,

0. cos 0 .d E d A EdA EdAφ = = =

• If we take any such element on the curved surface, area

vector d A

will be parallel to .E

Hence, flux due to all such elements on the Gaussian surface

.

...(1)

E d A EdA

E dA

φ

φ

= =

⇔ =

∫ ∫∫

• Now, dA =∫ surface area of Gaussian surface = 4𝜋𝜋r2 Hence, eqn (1) becomes

24E dA E rφ π= =∫ • Now we are ready to use Gauss’s Law, which is:

2enc

0 0

20

4

4

q QE r

QEr

φ πε ε

πε

= ⇔ =

⇔ =

• Clearly this expression for electric field due to a point charge is same as that of when derived by Coulomb’s Law.

EElleeccttrriicc ffiieelldd dduuee ttoo aann iinnffiinniittee lliinnee cchhaarrggee

• Consider a long line charge with a linear charge density λ. We have to calculate the electric field at a point P which is at a distance r from the line charge. The electric field must be along the radial outward direction.

• We draw a cylinder of length l passing through P and coaxial with the line charge as a Gaussian surface.

• The electric field at all the points on the curved surface have the same magnitude E as that at P. Also, the direction of the field at any point on the curved surface is normal to the line charge and hence normal to the cylindrical surface element there. The flux through the curved part is, therefore,

Flux through curved part:

0. cos 0 2 .E d S E dS E dS E dS E rlπ= = = =∫ ∫ ∫ ∫

• Now, consider the flat parts of the Gaussian surface. The field and the area-vector make an angle of 90o with each other so that

Flux through flat parts:

. 0E d S =∫

on these parts.

• Cleary, net total flux through the closed Gaussian surface Net flux through Gaussian surface:

( ). 2 . ... iE d S E rlφ π= =∫

• The charge enclosed in the Gaussian surface isGauss’s Law:

encq = λl as a length l of the line charge is inside the closed surface. Using gauss’s law and (i), we have,

( )enc

0 0

0

2

(Electric field due to a line charge) 2

lqE rl

Er

λφ π

ε ελ

πε

= ⇒ =

⇒ =

• This is the field at a distance r from the line. It is directed away from the line if the charge is positive and towards the line if the charge is negative.

• Variation of electric field with distance r from line charge is shown in given graph. Clearly, electric field E decreases with distance r from line charge.

r

Q d A

E

r E

dA

P

+ +

+ +

+

+ +

+

+ +

E

r

E



Electric Field due to a Uniformly Charged Spherical Shell

Suppose a total charge Q is uniformly distributed in a non-conducting spherical shell of radius R and we are required to find the electric field at a point P which is at a distance r from the centre of the charge distribution.

Field at an outside point:

• For a point P outside the charge distribution we have r > R. Draw a spherical Gaussian surface passing through the point P and concentric with the charge distribution.

• The electric field is radial by symmetry and if Q is positive the field is outside. Also, its magnitude E at all the points of the Gaussian surface must be equal. As the field E

is normal to the surface element everywhere, . E d S E dS=

for each element. The flux of the electric field through this closed surface is

2. 4 .E d S E dS E dS E rφ π= = = =∫ ∫ ∫

• Using Gauss’s Law, we get 2

0

20

4

.4

QE r

QEr

πε

πε

=

⇒ =

• The electric field due to a uniformly charged spherical shell at a point outside it, is identical with the field due to an equal point charge placed at the centre.

• Suppose, we wish to find the electric field at a point P inside the spherical charge distribution. We draw a spherical surface of radius r passing through P and concentric with the given charge distribution.

Field at an internal point:

• Clearly, the entire charge is on the surface of the spherical shell. Therefore, there is no charge inside our Gaussian surface, as it is inside the shell. Hence, enc 0.q =

• Using Gauss’s law,

enc

0

. 0

0

qE d S

E

φε

= ⇒ =

⇒ =

∫

• The electric field due to a uniformly charged spherical

shell at an internal point is zero. • Variation of electric field with distance r from the centre

is shown in following graph:

Electric Field due to a Uniformly Charged Solid Sphere

Suppose a total charge Q is uniformly distributed in a non-conducting spherical volume of radius R and we are required to find the electric field at a point P which is at a distance r from the centre of the charge distribution.

Field at an outside point:

• For a point P outside the charge distribution we have r > R. Draw a spherical Gaussian surface passing through the point P and concentric with the charge distribution.

• The electric field is radial by symmetry and if Q is positive the field is outside. Also, its magnitude E at all the points of the Gaussian surface must be equal. As the field E

is normal to the surface element everywhere, . E d S E dS=

for each element. The flux of the electric field through this closed surface is

2. 4 .E d S E dS E dS E rφ π= = = =∫ ∫ ∫

• Using Gauss’s Law, we get 2

0

20

4

.4

QE r

QEr

πε

πε

=

⇒ =

• The electric field due to a uniformly charged sphere at a point outside it, is identical with the field due to an equal point charge placed at the centre.

r R

S

S

E

P

E

E

E

E

E

E

dA

r

R

P

r

E

R

r R

S

S

E

P

E

E

E

E

E

E

dA



• Suppose, we wish to find the electric field at a point P inside the spherical charge distribution. We draw a spherical surface passing through P and concentric with the given charge distribution.

Field at an internal point:

• The radius of this sphere will be r. All the points of this sphere are equivalent. By symmetry, the field is radial at all the points of this surface and has a constant magnitude E. The flux through this spherical surface is

2

.

4 ...(1)

E d S E dS E dS

E r

φ

φ π

= = =

⇒ =

∫ ∫ ∫

• Let us now calculate the total charge contained inside this

spherical surface. As the charge is uniformly distributed within the given spherical volume, the charge per unit

volume is 3

.43

Q

Rπ The volume enclosed by the Gaussian

surface, through which the flux is 34 .3

rπ Hence, the

charge enclosed is 3

3enc 3

3

4. ...(ii)4 33

Q Qrq rRR

ππ

= =

• Using Gauss’s law, (i) & (ii), 3

2enc3

0 0

30

4

4

q QrE rR

QrER

φ πε ε

πε

= ⇒ =

⇒ =

• The electric field due to a uniformly charged sphere at an

internal point is 304

QrRπε

in radial direction.

• Variation of electric field with distance r from the centre is shown in following graph:

Electric Field due to a Plane Sheet of Charge

• Consider a large plane sheet of charge with uniform surface charge density σ. We have to find the electric field E at a point in front of the sheet.

• The cylinder together with its cross-sectional areas forms a closed surface and we apply Gauss’s law to this surface.

• The electric field at all the points on the right face has the same magnitude E. Thus, the flux of the electric field through right surface is

. AE A E=

• Similarly, the flux of the electric field through left face is

also E A. • At the points on the curved surface, the field and the

outward normal make an angle of 90o with each other and hence, flux . 0E A =∫

• The total flux through the complete closed surface is . 0E d S EA EAφ = = + +∫

2 EAφ⇒ = …(i)

• The area of the sheet enclosed in the cylinder is A. The charge contained in the cylinder is, therefore,

encq = σ A …(ii)

• Hence from Gauss’s law,

enc

0 0

2q AEA σφε ε

= ⇒ =

0

2

E σε

⇒ =

• We see that the field is uniform and does not depend on the distance from the charge sheet. This is true as long as the sheet is large as compared to its distance from P.

E

r

R

dA

P

r

E

R

P

E

+

+

+

+

+

+

+

+

+ E

A +

+

d S

+

+

+ +

+

+ +

+

+ A

E



Electric Field near a Charged Conducting Surface

• Let us consider a large, plane conducting sheet. The surface on right has a uniform surface charge density σ.

• We have to find the electric field at a point P near this surface and outside the conductor. As we know, the conducting surface is an equipotential surface and the electric field near the surface is perpendicular to the surface. For positive charge on the surface, the field is away from the surface.

• To find the electric field, we construct a Gaussian surface

as shown in figure. We draw a cylinder whose left side is terminated in the interior of the conducting sheet.

• The electric field at all the points on the right face has the same magnitude E. Thus, the flux of the electric field through right surface is

.E A EA=

• At the points on the curved surface, the field and the

outward normal make an angle of 90o with each other and hence, flux . 0E A =∫

• Also, the flux on the left face is zero as the field inside the conductor is zero in electrostatic.

• The total flux through the Gaussian surface constructed is, therefore, EAφ = …(i)

• The charge enclosed inside the closed surface is

encq = σ A …(ii) • Hence from Gauss’s law,

enc

0 0

q AEA σφε ε

= ⇒ =

0

E σε

⇒ =

• The electric field near a charged conducting surface is

0

σε

and it is normal to the surface.

Quick recap

• Area vector of a flat surface is a vector perpendicular to surface having magnitude equal to its area A. Area vector is taken in outward normal direction for a closed surface.

• Electric flux: The total number of electric field lines passing normally through an area placed inside an electric field is called electric flux.

.E Aφ =

(Electric flux through a flat area in a uniform field)

E d Aφ = ⋅∫ (Electric flux for general case)

E d Aφ = ⋅∫

(Electric flux through a closed surface)

• Gauss’s Law: Gauss's law states that the total electric flux

through any closed surface is 0

1ε

times total charge enclosed

by that surface. Thus,

( )

( )

enc

0

enc

0

Gauss's Law

. Gauss's Law

q

qE d A

φε

ε

=

=∫

Important points to note about Gauss’s Law: 1. φ = electric flux because of electric field created by all

charges (inside and outside the Gaussian surface) 2. q

enc = total charges which are inside the Gaussian surface.

• Gauss’s Theorem for any medium: enc enc

0 0

r

q qφε ε ε

= =

• Flux density = Total fluxArea

• Electric field due to a infinite long charge: 0

2

Er

λπε

=

• Electric field due to a non-conducting infinite plane sheet

of charge: 0

2

E σε

=

• Electric field near a conducting surface: 0

E σε

=

• Electric Field due to a uniformly charged Spherical Shell

Inside: 0E = ; Outside:

20

.4

QErπε

=

• Electric Field due to a uniformly charged non-conducting solid sphere

Inside: 30

; 4

QrERπε

= Outside:

20

.4

QErπε

=

• Electric Field due to a conducting solid sphere

Inside: 0E = ; Outside:

20

.4

QErπε

=

A

E

+

+

+

+

+

+

+

+

+ E

+

+

n

+

+

+ +

+

+ +

+

+ E=0

P

A



Example 2

Find the flux of the electric field through a spherical surface of radius R due to a charge of 10-7 C at the centre and another equal charge at a point 2R away from the centre. Solution: • Here only one charge is inside the sphere. Hence, from

Gauss’s Law, flux through sphere

enc

0 07

12

4 2 1

108.85 101.1 10 Nm C

q Qφε ε

φ

φ

−

−

−

= =

⇔ =×

⇔ ≈ ×

Example 3

Q. Show that there can be no net charge in a region in which the field is uniform at all points. Solution: • We know that the net flux through the surface is

proportional to the net number of lines leaving the surface, where the net number means the number leaving the surface minus the number entering the surface.

• Uniform electric field means that electric field lines are parallel and spacing between any two field line is same.

• Hence, if a closed surface is placed in a region of uniform electric field, the net number of lines leaving the surface is zero, since same number of field lines leave as that which enter.

• Thus, flux ø through a closed surface in a uniform electric field is zero.

• Hence, from Gauss’s Law

enc

0

enc

0

enc

0

0.

q

q

q

φε

ε

=

⇔ =

⇔ =

Hence, there can be no net charge in a region in which the field is uniform at all points.

Example 4

A charge Q is placed at a distance 2a above the centre of a

horizontal, square surface of edge a as shown in figure. Find the flux of the electric field through the square surface.

Solution: • This problem is bit tricky. The key idea is that we can

find flux using .E d Aφ = ∫

but that process will be quite

cumbersome. • Another key idea is that we can use Gauss’s Law to find

the flux; we need a closed surface for that. One important concept is that we should use Gauss’s Law for the case when the situation is symmetrical. So, what to do?

• As the square surface has edge a and also the charge Q is

kept at a distance 2a from the centre of the surface, we

can construct a cube of edge a having one face coincident with given surface, such that it encloses charge Q at the centre.

• Hence, from Gauss’s Law, flux through the entire cube

enc

0 0

q Qφε ε

= =

• As the situation is symmetrical each face will share same

amount of flux. • Hence, flux through required surface

0

Total flux6 6

Qε

= =

Quick Exercise 3

1. Four closed surfaces, S1 through S

4 , together with the

charges -2Q, Q, and -Q are sketched in Figure. Find the electric flux through each surface.

2. Figure shows six point charges that all lie in the same plane. Five Gaussian S

1 through S

5, each enclose part of

this plane, and Fig. shows the intersection of each surface with the plane. Rank these five surfaces in order of the electric flux through them, from most positive to most negative.

2R Q

Q

a

a

/ 2a Q

Q



Example 4

• Q. A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r1 & r2 respectively. Find the electric field at a point P a distance x away from the centre for r1 < x < r2. Draw a rough diagram showing the electric field as a function of x for 0 < x < 2r2

Solution: • This problem is a simple application of Gauss’s Law. • Let us consider a Gaussian spherical surface at of radius x

concentric with given sphere. • To use Gauss’s law we must have to find out the total

charge inside the Gaussian surface. • As the given sphere is hollow, its volume is given by :

( )3 32 14

3

r rV

π −=

• Therefore, charge density

( )3 32 1

34

Q QV r rπ

= =−

• Hence, charge inside the Gaussian surface q

enc = (charge density) × (volume of shell with inner and

outer radii r1 & x)

( )( )

( )( )

( )3 3 3 3

1 1enc 3 3 3 3 3 3

2 1 2 1 2 1

43 ...(1)4

r r Q r rQqr r r r r r

π

π

− −⇔ = × =

− − −

• Hence, using Gauss’s Law, flux through Gaussian

surface:

( )( )

3 31enc

3 30 0 2 1

...(2)Q x rq

r rφ

ε ε

−= =

−

• As the situation is symmetrical, field at any point must be along radial direction. Since, area vector is in the outward normal direction, area vector d S

at each point of

Gaussian surface will be along .E

• Hence, flux through Gaussian surface is

0. cos 0E d S EdS EdSφ = = =∫ ∫ ∫

(because angle between 0& 0E d S =

)

E dSφ⇔ = ∫

(Since magnitude of field is constant at each point of

Gaussian Surface)

Φ = E S = E (4𝜋𝜋 x2)

…(3)

(surface area of Gaussian surface = 4 𝜋𝜋 x2) • Hence, using (2) & (3), we get

( ) ( )( )

( )( )

3 312

3 30 2 1

3 31

2 3 30 2 1

4

...(4)4

Q x rE x

r r

Q x rE

x r r

πε

πε

−=

−

−⇔ =

−

•

• For region x < r1: Sphere is hollow, therefore there can’t be any charge inside it. Hence, there can’t be any electric field inside the region x < r1 . Plot is roughly shown (blue colored)

Plotting the electric field as a function of x for 0 < x < 2r2

• For region r1 < x < r2 : Electric field is given by

expression (4) of the last slide. Thus, electric field increase here. Plot is toughly shown (red colored).

• For region x > r2: Outside region of sphere. Here, spherical shell charge behaves as if the net charge is placed at the centre. Thus field at a distance x from centre outside the sphere is given by:

20

.4

QExπε

=

Thus field decreases with distance which can be plotted roughly (sky blue).



Exercises

NCERT Problems 1.17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8 × 103 Nm2 C-1. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not? 1.18. A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm. What is the magnitude of the electric flux through the square? 1.19. A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface? 1.20. A point charge causes an electric flux of –1.0× 103 Nm2 C-1

to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge? 1.21. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere? 1.22. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? 1.23. An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density. 1.24. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17 × 10–22 C/m2. What is E? (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? 1.28. (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

1.29. A hollow charged conductor has a tiny hole cut into its

surface. Show that the electric field in the hole is

0

n,2

E σε

=

where n is the unit vector in the outward normal direction, and σ is the surface charge density near the hole. 1.30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law.

More problems for practice:

1. Consider Gauss law 0

. ,qE dsε

=∫ then for the

situation shown in the figure at through Gaussian surface, (a) the E due to q2 would be

zero (b) the E due to both q1 and

q2 would be non-zero (c) the φ due to q1 and q2

would be on-zero (d) the φ due to q2 would be

zero

2. In a region of the space, the electric field is in the x-

direction and proportional to 0, . ., .x i e E E xi=

Consider an imaginary cubical volume of edge a, with its edges parallel to the axes of coordinates. The charge inside this volume is (a) zero (b) ε0E0a3 (c) 3

0 0E aε

(d) 20 0

16

E aε

3. A charge Q is placed at the centre of the mouth of a

conical flask. The flux of the electric field through the flask is (a) Zero

(b) 0

Qε

(c) 02

Qε

(d) 02

Qε

<

4. A long string with a charge of λ per unit length

passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be

(a) 0

aλε

(b) 0

2 aλε

(c) 2

0

6 aλε

(d) 0

3 aλε

q1

q2



5. Charges Q1 and Q2 lie inside and outside respectively

of a closed surfaces S. Let E be the field at any point on S and φ be the flux of E over S. (a) If Q1 changes, both E and φ will change. (b) If Q2 changes, E will change but φ will not

change. (c) If Q1 = 0 and Q2 ≠ 0 then E ≠ 0 but φ = 0. (d) If Q1 ≠ 0 and Q2 = 0 then E = 0 but φ ≠ 0.

6. Charges Q1 and Q2 are placed inside and outside

respectively of an unchanged conducting shell. Their separation is r. (a) The force on Q1 is zero.

(b) The force on Q1 is 1 22 .Q Qk

r

(c) The force on Q2 is 1 22 .Q Qk

r

(d) The force on Q2 is zero.

7. A spherical conductor A lies inside a hollow spherical conductor B. Charges Q1 and Q2 are given to A and B respectively. (a) Charge Q1 will appear on the outer surface of A. (b) Charge –Q1 will appear on the inner surface of

B. (c) Charge Q2 will appear on the outer surface of B. (d) Charge Q1 + Q2 will appear on the outer surface

of B.

8. A disk of radius a / 4 having a uniformly distributed charge 6 C is placed in the x–y plane with its centre at (–a / 2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is place on the x-axis from x = a / 4 to x = 5a / 4. Two point charges –7 C and 3 C are placed at (a / 4, – a / 4, 0) and (–3a/4, 3a / 4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a / 2, y = ± a / 2, z = ± a / 2. The electric flux through this cubical surface is (IIT-JEE 2009) (a) 02 /C ε−

(b) 02 /C ε

(c) 010 /C ε

(d) 012 /C ε

9. A cubical region of side a has its centre at the origin.

It encloses three fixed point charges, − q at (0, − a/4, 0), + 3q at (0, 0, 0) and −q at (0, +a/4, 0). Choose the correct option(s). (IIT-JEE 2012)

a) The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane

x = −a/2. b) The net electric flux crossing the plane y = +a/2

is more than the net electric flux crossing the plane y = −a/2.

c) The net electric flux crossing the entire region

is 0

qε

d) The net electric flux crossing the plane z = +a/2

is equal to the net electric flux crossing the

plane x = +a/2.

10. A charge Q is placed at the corner of a cube. The flux of the electric field through the cube is (a) Zero

(b) 0

Qε

(c) 02

Qε

(d) 08

Qε



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