Chapter 6 Electrostatic Boundary Value

Problems

Dr. Talal Skaik

2012 1

Introduction • In previous chapters, E was determined by coulombs law or Gauss

law when charge distribution is known, or when

potential V is known throughout the region.

• In most practical applications, however, neither the charge

distribution nor the potential distribution is known.

• In this chapter, practical electrostatic problems are considered

where only electrostatic conditions (charge & potential) at some

boundaries are known, and E and V are to be found.

• Such problems are tackled using Poisson's or Laplace’s equation

or the method of images.

2

E V

6.2 Poisson’s and Laplace’s Equations

3

• They are easily derived from Gauss’s law.

2

2

poisson's e

D

quation

Lapl

E

E

for an inhomogeneous medium

for homogeneous medium ( )

when 0, (for charge free region):

0 ( )ace's equation

V

V

V

V

V

V

V

V

2 2 2

2 2 2

2 2

2 2 2

22

2 2 2 2 2

0

1 10

1 1 1sin 0

sin sin

V V V

x y z

V V V

z

Cartesian

Cylindrical

Spherical

V V Vr

r r r r r

Laplace’s Equation

4

2 0V

2 2 2

2 2 2

2 2

2 2 2

22

2 2 2 2 2

1 1

1 1 1sin

sin sin

V

V

V

V V V

x y z

V V V

z

V V Vr

r r r r r

Cartesian

Cylindrical

Spherical

5

Poisson’s Equation

2 VV

Uniqueness Theorem • There are several methods of solving a given problem (analytical,

graphical, numerical, experimental, etc).

• The question is: If a solution of Laplace’s equation satisfies a given

set of boundary conditions, is this the only possible solution?

• The answer is Yes, there is only one solution, and it is unique.

• This is uniqueness theorem: If a solution of Laplace’s equation

can be found that satisfies the boundary conditions, then the

solution is unique.

6

General Procedure for Solving Poisson's or Laplace's Equation

7

1. Solve Laplace's (if ) or Poisson's ( if ) equation using:

(a) direct integration when V is a function of one variable, or

(b) separation of variables if V is a function of more than one

variable.

The solution at this point is not unique but expressed in terms of

unknown integration constants to be determined.

2. Apply the boundary conditions to determine a unique solution for V.

Imposing the given boundary conditions makes the solution unique.

3. Having obtained V, find E using and D using . E V

0V 0V

D E

Current-carrying components in high-voltage power equipment must

be cooled to carry away the heat caused by ohmic losses. A means of

pumping is based on the force transmitted to the cooling fluid by

charges in an electric field. The electrohydrodynamic (EHD) pumping

is modeled in the Figure. The region between the electrodes contains

a uniform charge ρ0 which is generated at the left electrode and

collected at the right electrode. Calculate the pressure of the pump if

ρ0 =25 mC/m3 C and V0 =22 kV.

8

Example 6.1

•The technology promises to make it easier and more efficient to

remove heat from small spaces, vastly expanding the capabilities of

advanced instruments and microprocessors.

•Unlike traditional thermal-control technologies that rely on

mechanical pumps and other moving parts, EHD cooling uses

electric fields to pump coolant through tube.

•Electrodes apply the voltage that pushes the coolant through tube.

•EHD-based systems are lightweight and consume little power. 9

EHD pumps

V

2

0

2 2 2

2 2 2 2

Since 0, we apply poisson's equation

the boundary conditions V(z=0)=V and V(z=d)=0 show that V depends only on z,

1 1Hence becomes:

Integratin

V

V o

V

V V V V

z z

2

0 0 0

g once gives

Integrating again gives =2

where A and B are integration constants to be determined by applying

the boundary conditions.

when 0, =0 0

when ,

o

o

dV zA

dz

zV Az B

z V V V B B V

z d

2

000 0= or A=

2 2

o od d VV Ad V

d

10

Example 6.1 - Solution

0

0

0

200

0

The electric field is given by

1E= V=

E2

The net force is F= E,

F= E E

2

z

o oz z z

T T v

v

d

v

z

d

oz

V V Va a a

z

z VV da A a z a

z d

Q Q dv

dv dS dz

V zS z dz a

d

0 0

3 3 2

0 0

F= a

Force per unit area (Pressure) is 25 10 22 10 550 N/m

zSV

FSV

S

11

Example 6.1 - Continued

2

0

0

=2

=2

o

o

zV Az B

B V

d VA

d

12

Example 6.2 The xerographic copying machine is an important application of

electrostatics. The surface of the photoconductor is initially charged

uniformly as in Figure 6.2(a). When light from the document to be

copied is focused on the photoconductor, the charges on the lower

surface combine with those on the upper surface to neutralize each

other. The image is developed by pouring a charged black powder

over the surface of the photoconductor. The electric field attracts

the charged powder, which is later transferred to paper and melted

to form a permanent image. We want to determine the electric

field below and above the surface of the photoconductor.

13

Example 6.2

14

Example 6.2

V

22

2

1 2

1 1 1

Since 0, we apply Laplace's equation.

Also potential depends only on x.

0

Integrate twice gives

V=Ax+B

Let the potentials above x=a be V , and below x=a be V .

V =A x+B x>a

V

VV

x

2 2 2

1 1 1 1 1

2 2 2

=A x+B x<a

The boundary conditions at the grounded electrodes are,

V (x=d)=0 0=A d+B B = A d (1)

V (x=0)=0 0=0+B B =0 (2)

15

Example 6.2 - Solution

1 2 1 1 2

1n 2n 1 1n 2 2n

1 21 2 1 1 2 2

At the surface of the photoconductor,

V (x=a)=V (x=a) A a+B A a (3)

=

Since E= V=

Therefore (4)

S Sx a

x y z n

S S

D D E E

dV dV dV dVa a a E

dx dy dz dx

dV dVA A

dx dx

11 1

2 21

1 1

22 2

2 21

1 1

Solve equations (1) to (4) to find the constants.

aE = a a ,

1

1 a

E = a a , 0

1

S xx x

S x

x x

dVA a x d

dx d

a

d

dV aA x a

dx d

a

16

Example 6.2 - Continued

1 1 1

2 2 2

1 1 2

V =A x+B x>a

V =A x+B x<a

B = A d, B =0

17

Example 6.3 Semi-infinite conducting planes at Ф=0 and Ф=π/6 are separated by an infinitesimal insulating gap as in Figure6.3. If V(Ф=0)=0 and (Ф=π/6 )=100 V, Calculate V and E in the region between the planes.

18

Example 6.3 Solution

2 2 22 2

2 2 2 2 2

Since depends only on , Laplace's equation in cylindrical coordinates becomes

1 1 10 0 0

since =0 is excluded due to the insulating gap, we can multip

V

V V V d VV V

z d

22

2

0 00 0 0 0

0 0

0

0

0

ly by to get 0

Integrating twice gives V=A +B.

Apply boundary conditions:

when 0, 0 0=0+B B=0

when , V =A , Hence =

1

Substituting V =100 an

d V

d

V

V VV V A V

VdVE V a a

d

0

2

600 600d / 6 gives: = and E=

check V( 0)=0, V( / 6)=100, 0

V a

V

19

Example 6.4 Two conducting cones (Ѳ=π/10 and Ѳ=π/6) of infinite extent are separated by an infinitesimal gap at r = 0. If V(Ѳ=π/10 )=0 and V(Ѳ=π/6)=50 V, Find V and E between the cones.

2

2

2

Since depends only on , Laplace's equation in spherical

1coordinates becomes: sin 0

sin

since =0 and =0, are excluded, we can multiply by sin to get

s

V

d dVV

r d d

r r

d

d

2

in 0

Integrating once gives: sin sin

Integrating this results in =sin 2cos( / 2)sin( / 2)

cos( / 2) 1 sec ( / 2)

2cos( / 2)sin( / 2) cos( / 2) 2 tan( / 2)

Let = tan

dV

d

dV dV AA

d d

d dV A A

d dV A A

u

2/ 2, =1/ 2sec ( / 2)

= ln( ) ln(tan / 2)

du d

duV A A u B V A B

u

20

Example 6.4 Solution

1 1 1

1

1

2 0 0

Apply boundary conditions to determine the integration constatnts and .

( ) 0 0 ln(tan / 2) ln(tan / 2)

tan / 2Hence ln(tan / 2) ln(tan / 2) ln

tan / 2

Also ( ) l

A B

V A B B A

V A A V A

V V V A

02

1 2

1

0

1

2

1

tan / 2n

tan / 2 tan / 2ln

tan / 2

tan / 2ln

tan / 2Thus =

tan / 2ln

tan / 2

VA

V

V

21

Example 6.4 Continued ln(tan / 2)V A B

22

Example 6.4 Continued

0

2

1

1 2 0

1

sin tan / 2sin ln

tan / 2

Taking , , and 50 gives10 6

tan / 250ln

tan / 2tan / 20= 95.1ln

tan /12 0.1584ln

tan / 20

and

95.1E=

sin

VdV AE V a a a

r d rr

V

V V

r

2

0

V/m

Check: V( /10)=0, V( / 6)=V , 0

a

V

1

0

2

1

0

1

2

1

tan( / 2)ln

tan( / 2)

tan( / 2)ln

tan( / 2)

tan( / 2)ln

tan / 2=

tan( / 2)ln

tan( / 2)

V A

VA

V

V

sin

dV A

d

6.5 Resistance and Capacitance

23

• For conductor with uniform cross section, the resistance is

• If the cross section of the conductor is not uniform, the resistance is obtained from:

lR

S

E. l

E. S

dVR

I d

24

The resistance R ( or conductance G=1/R) of a conducting material

can be found by the following steps:

1. Choose a suitable coordinate system.

2. Assume Vo as the potential difference between conductor

terminals.

3. Solve Laplace's equation to obtain V then determine E

from and find I from .

4. Finally, obtain R as Vo/I.

2 0V

E V E. SI d

Resistance

Capacitance • To have capacitance, we must have two (or more) conductors

carrying equal but opposite charges.

• The conductors may be separated by free space or dielectric.

• The conductors are maintained at a potential difference:

• (note E is always normal to conducting surfaces)

25

1

2

E. lV d

The capacitance C of the capacitor is defined as the ratio of the

magnitude of the charge on one of the plates to the potential

difference between them:

Obtain C for any two conductor capacitors by either:

1. Assume Q and determine V in terms of Q (involves Gauss Law).

2. Assume V and determine Q in terms of V (involves Laplace’s equation)

Method (1) involves the following:

1. Choose a suitable coordinate system.

2. Let the two conducting plates carry charges +Q and -Q.

3. Determine E using Coulomb or Gauss law and find V from

4. Finally, obtain C from C=Q/V. 26

E. S

E. l

dQC

V d

E. lV d

Capacitance

Storing Charges- Capacitors: A capacitor consists of 2 conductors of any shape placed near one another without touching. It is common; to fill up the region between these 2 conductors with an insulating material called a dielectric. We charge these plates with opposing charges to set up an electric field. 27

Capacitors

Capacitance or Capacitance Resistance

28

1

2

E l

E Gauss or

Coulombs law

(in terms of Q)

QC

V

V d

0

2

2

D S

D= E

E=

=0

or = /V

QC

V

Q d

V

V

V

0

2

2

= J S

J= E

E=

=0

or = /V

VR

I

I d

V

V

V

Suppose each plate has an area S and the plates are separated by distance d. Assume plates 1 and 2 carry uniformly distributed charges +Q and –Q 29

Parallel Plate Capacitor

1

2 0

22

E

, or ( )

Hence E l

and thus for a parallel-plate capacitor

1 1The energy stored in a capacitor is W

2 2 2

SS S x x x

d

x x

Q QD a E a a

S S

Q QdV d a dx a

S S

Q SC

V d

QCV QV

C

Assume Conductors 1 and 2 carry uniformly distributed charges +Q and –Q. Apply gauss law to arbitrary Gaussian cylindrical surface of radius ρ such that a< ρ< b, 30

Coaxial Capacitor

1

2

E S= 2 E=2

E l ln2 2

Thus the capacitance of a coaxial cylinder is given by

2C=

ln

a

b

QQ d E L a

L

Q Q bV d a d a

L L a

Q L

bV

a

Assume charges +Q and –Q are on inner and outer spheres. Apply gauss law to arbitrary Gaussian spherical surface of radius r such that a< r< b, 31

Spherical Capacitor

2

2

1

2

2

E. S= 4 E=4

1 1E l a a

4 4

Thus the capacitance of a spherical capacitor is given by

4 C=

1 1

r r

a

r r

b

QQ d E r a

r

Q QV d dr

r a b

Q

V

a b

32

Series and Parallel Capacitors •If two capacitors with capacitance C1 and C2

are in series ( have same charge on them), the

total capacitance is:

•If the capacitors are in Parallel (have same voltage

across them), the total capacitance is:

1 2

1 2 1 2

1 1 1 or C=

C C

C C C C C

1 2C C C

It has been shown that:

Which is the relaxation time Tr of the medium separating the conductors.

33

E l

E S

E S

E l

The product of these expressions yields

dVR

I d

dQC

V d

RC

Note: R in these equations is NOT the resistance of the capacitor plate, but the leakage resistance between the plates. Therefore, σ is the conductivity of the dielectric medium separating the plates. 34

Assuming homogeneous media, the resistance of various

capacitors mentioned earlier can be obtained using RC = ε/σ :

For a parallel-plate capacitor ,

ln2

For a cylindrical capacitor , 2ln

1 14

For a spherical capacitor , 1 1 4

S dC R

d S

bL aC R

b L

a

a bC R

a b

35

Example 6.8

A metal bar of conductivity σ is bent to form a flat 90o sector of inner

radius a, outer radius b, and thickness t as shown in the Figure. Show

that :-

(a) The resistance of the bar between the vertical curved surfaces at r = a

and r = b is:

(b) The resistance between

the two horizontal surfaces

at z=0 and z=t is

2lnb

aRt

2 2

4'

( )

tR

b a

0

(a) Between the vertical curved ends located at =a and =b, the bar has

a nonuniform cross section. Let a potential difference V be maintained

between the curved surfaces at = and =b so that V(a

0

2

= )=0 and

V( = )=V . Solve V using Laplace's equation in cylindrical coordinates,

1since V=V( ), 0

As 0 is excluded, multiply by and

integrate once, , i

a

b

d dVV

d d

dV dV AA

d d

0 0

ntegrating again yields,

V=A ln +B , where A and B are constants of integration.

Use the boundary conditions:

( = )=0 0=A ln +B or ln

( = )= =A ln +B=A ln ln ln or

V a a B A a

bV b V V b b A a A A

a

0

ln

V

b

a36

Example 6.8 Solution

0

0

/2

0 0

0 0

0

Hence,

ln ln a= ln ln

ln

ln

,

2

ln ln

2ln

Thus R=

t

z

VV A A A

ba a

a

VdV AE V a a a

bd

a

J E dS d dza

V tVI J dS dz d

b b

a a

bV

I

as requireda

t37

Example 6.8 Continued

0

V=A ln +B

ln

ln

VA

b

a

B A a

0

0

2

2

(b)

Let be the ptential difference between the two horizontal surfaces

so that V( 0)=0 and V( )=V .

since ( ),Laplace's equation becomes

0, integrating twice gives

, apply bou

V

z z t

V V z

d V

dz

V Az B

00 0

0

0

ndary conditions to get and ,

V( 0)=0 0 0 0

V( )= A

Hence,

=z z

A B

z B or B

Vz t V V At or

t

VV z

t

VdVE V a a

dz t

38

Example 6.8 Continued

39

Example 6.8 Continued

0

/2 2 22

0 0 0

0

0

2 2

J ,

( )J .

2 2 4

4Thus R'=

Alternatively, for this case, the cross section of the bar is uniform

between the horizo

z z

bb

a a

VE a dS d d a

t

V V V b aI dS d d

t t t

V t

I b a

2 22 2

ntal surfaces at z=0 and z=t, then

4R'=

( )4

l t t

S b ab a

A coaxial cable contains an insulating material of conductivity σ . If the radius of the central wire is a and that of the sheath is b, show that the conductance of the cable per unit length is:

40

Example 6.9

2G=

lnb

a

Transmission Line Equivalent circuit: (See Chapter 11)

41

Example 6.9

2G=

lnb

a

G=Conductance per unit length for dielectric.

G=1/(Leakage Resistance of Dielectric)

0

0

2

0

Let V be the potential difference between the inner and outer conductors, so that

( ) 0 and ( )

1Solve 0. From example 6.8, we got:

ln

ln

V a V b V

d dVV

d d

VV

b a

a

0

0

2

0 0

0 0

0

and =

ln

,

ln

2

ln ln

ln1

The resistance per unit length is R= .2

1 2and the conductance per unit length is G=

ln

l

z

VE a

b

a

VJ E a dS d dza

b

a

V L VI J dS dzd

b b

a a

bV a

I L

R

b

a42

Example 6.9 Solution

Conducting spherical shells with radii a = 10 cm and b = 30 cm are maintained at a potential difference of 100 V such that V(r=b) = 0 and V(r=a) = 100 V. Determine V and E in the region between the shells. if εr =2.5 in the region, determine the total charge induced on the shells and the capacitance of the capacitor. 43

Example 6.10

2 2

2

2

2 2

2

10

since 0 in the region of interest, we multiply through by to obtain

0, Integrating once gives: or

Integrating again: (use bo

d dVV r

r dr dr

r r

d dV dV dV Ar r A

dr dr dr dr r

AV B

r

00 0

0

undary conditions to find A and B)

1 1when , 0 0= or B= , Hence

1 1when , , or A=

1 1

1 1

Thus =1 1

A Ar b V B V A

b b b r

Vr a V V V A

b a

b a

r bV V

a b

44

Example 6.10 Solution

0

22

2

20 0 0 0

20 0

2

E1 1

To determine capacitance, first find Q on either shell

4E sin

1 11 1

, Q= For inner shell, Q= [4 ], bu

r r r

r r

S S

VdV AV a a a

dr rr

a b

V VQ dS r d d

ra ba b

Or dS a

20 0 0 0 0 0

2 2

0

t

4 Q= .4

1 1 1 1 1 1

The capacitance is easily determined as

4C=

1 1

S n r a

r r rn nr a r a

D

V V VD E a

a aa b a b a b

Q

V

a b

45

Example 6.10 Continued

0

0

1 1

A=1 1

1 1

=1 1

V Ab r

V

b a

r bV V

a b

0

2 2

9

Substituting =0.1 m, =0.3 m, =100 V yields,

1 10

1 103=100 15

10 10 / 3 3

100 15 V/m

10 10 / 3

104 (2.5)(100)

36Q= 4.167 nC

10 10 / 3

positive charge is induced on the inner s

r r

a b V

rV V

r

E a ar r

9

0

hell, negative charge

is induced on the outer shell.

4.167 10C= 41.67 pF

100

Q

V

46

Example 6.10 Continued 0

1 1

=1 1

r bV V

a b

In Section 6.5, it was mentioned that the capacitance C=Q/V of a capacitor can be found by either assuming Q and finding V or by assuming V and finding Q. The former approach was used in Section 6.5 while we have used the latter method in the last example. Using the latter method, derive capacitance for parallel plate capacitor:

0

0

Assuming the potentail difference between the paralle plates is

so that (x=0)=0 and (x=d)= .

V

V V V 47

Example 6.11

Q SC

V d

22

2

00 0 0

Solve Laplace's equation: 0

Integreating twice gives

At 0, 0 0 0 or 0

At , 0 or / . Hence =

To find the capacitance, first find the charge o

d VV

dx

V Ax b

x V B B

Vx d V V V Ad A V d V x

d

0n

0 0

0 0

n either plate

But D = E a , where E

On the lower plate , So and

On the upper plate , So and

(Q is equal

S

S n x x x

n x S

n x S

Q dS

dV Va V a Aa a

dx d

V V Sa a Q

d d

V V Sa a Q

d d

0

but opposite on each plate). C=Q S

V d

48

Example 6.11 Solution

Determine the capacitance of each of the capacitors in the Figure. Take εr1 =4, εr2=6, d=5 mm, S=30 Cm2.

1 2

0 1 0 1 0 21 2

(a) Since D and E are normal to the dielectric interface, the capacitor in

Fig. (a) can be treated as two capacitors C and C in series.

2 2= , =

/ 2

r r rS S SC C

d d d

49

Example 6.12

Example 6.12 Solution

50

9 4

1 2 0 1 2

3

1 2 1 2

The total capacitor is given by

2 ( ) 10 30 10 4 6= 2. . . =25.46 pF

36 5 10 10

(b)

D and E are parallel to the dielectric interface, the capacitor in

Fig. (b) can be treated a

r r

r r

C C SC C

C C d

1 2

1 2

0 1 0 1 0 21 2

9 4

01 2 1 2 3

s two capacitors and in parallel

(the same voltage across and )

/ 2= , C =

2 2

The total capacitance is

10 30 10= ( ) . .10 26.53 pF

2 36 (5 10 )

r r r

r r

C C

C C

S S SC

d d d

SC C C

d

Example 6.12 Solution Continued

0 0

0 0

E S= 2 . Hence E=2

2 2 2 1010

10ln(10 ) ln

2 2 10

Thus the capacitance per meter is (L=1m)

a a a

o rb b b

a

b

QQ d E L a

L

Q Q d Q dV d

L L L

Q Q bV

L L a

9

02 10 1C= 2 . . 434.6 pF/m

10 12.536ln ln

10 11.0

Q

bV

a

51

Example 6.13 A cylindrical capacitor has radii a = 1 cm and b = 2.5 cm. If the space between the plates is filled with an inhomogeneous dielectric with εr=(10+ρ)/ρ, where ρ is in centimeters, find the capacitance per meter of the capacitor.