Chapter 4: Solutions of Electrostatic Problems

4-1 Introduction

4-2 Poisson’s and Laplace’s Equations

4-3 Uniqueness of Electrostatic Solutions

4-4 Methods of Images

4-5 Boundary-Value Problems in Cartesian Coordinates

4-6 Boundary-Value Problems in Cylindrical Coordinates

4-7 Boundary-Value Problems in Spherical Coordinates

4.2 Poisson’s and Laplace’s Equations

Two fundamental equations for electrostatics

In a linear and isotropic medium

For a homogeneous medium and Poisson’s equation:

2 =V

-

and

（ ）=-Vand

2 22

2 2 2

1 1( )

V V VV r

r r r r z

Cylindrical coordinates:

22 2

2 2 2 2 2

1 1 1( ) (sin )

sin sin

V V VV R

R R R R R

Spherical coordinates:

2 0V Laplace’s equation (No free charges):

2 2 2

2 2 2=

V V V

x y z

-Cartesian coordinates:

Charges on the conductor surfaces

Electric field

Once V is known,

Q: Determine the field both inside and outside a spherical cloud of electrons with a uniform volume charge density ρ=˗ρ0 (where ρ0 is a positive quantity) for 0 ≤ R ≤ b and ρ=0 for R>b by solving Poisson’s and Laplace’s equation for V.

Q: Determine the field both inside and outside a spherical cloud of electrons with a uniform volume charge density ρ=˗ρ0 (where ρ0 is a positive quantity) for 0 ≤ R ≤ b and ρ=0 for R>b by solving Poisson’s and Laplace’s equation for V.

0 R b 0

2 =V

-

2 02

0

1( )idVdR

R dR dR

0 12

0

.3

idV CR

dR R

Electric field inside:

(1) Inside the cloud

Poisson’s equation

ρ=0 ρ=-ρ0

b

Ei cannot be infinite at R=0 C1=0

0

2 =0V

22

1( ) 0odVdR

R dR dR

Electric field outside:

(2) Outside the cloud

Laplace’s equation

0 22

dV C

dR R

Electric field continuity at R=b

30

4

3Q b

ρ=0 ρ=-ρ0

b

ρ=0 ρ=-ρ0

b

Potential inside:

Potential outside:

At boundary R = b, Vi = V0, we have

Calculate electric field from potential

Lastly, Vi Ei and Vo Eo.

4.3 Uniqueness of Electrostatic Solutions

Uniqueness theorem

A solution of Possion’s equation (of Laplace’s equation is a special case) that satisfies the given boundary conditions is a unique solution.

A solution of an electrostatic problem satisfying its boundary conditions is the only possible solution, irrespective of the method by with the solution is obtained.

4.4 Methods of Images

Methods of Images: replace boundary surfaces by appropriate image charges in lieu of a formal solution of Poisson’s or Laplace’s equation.

Grounded conducting plane

Grounded conducting plane

+

4.4-1 Point charge and conducting planes

Q: Consider the case of a positive point charge, Q , located at a distance d above a large grounded (zero-potential) conducting planeFind the potential at every point above the conducting plane (y>0).

The V(x,y,z) should satisfy the following conditions:(1) At all points on the grounded conducting plane, V(x,0,z)=0.(2) At points very close to Q, V approaches that of the point charge alone; that is as R→0

04

QV

R

(3) At points very far from , , V →0.(4) V is even with respect to the x and z coordinates:

V(x, y, z) = V(-x, y, z) and V(x, y, z) = V(x, y, -z)

( , , )Q x y z

Positive charge Q at y=d would induce negative charges on the surface of the conducting plane, resulting in a surface charge density ρs. Hence the potential at points above the conducting plane would be:

2 2 20 10

1( , , ) ,

44 ( )s

s

QV x y z ds

Rx y d z

Trouble: how to determine the surface charge density ρs?

If we remove the conductor and replace it by an image point charge -Q at y=-d, then the potential at a point P(x,y,z) in the y>0 region is

Note: The image charge should be located outside the region where the field is to be determined

12 2 2 2[ ( ) ]R x y d z

12 2 2 2[ ( ) ]R x y d z

Original charge

Image charge

Point charge and perpendicular conducting planes

Q: Consider a line charge ρl (C/m) located at a distance d from the axis of a parallel, conducting, circular cylinder of radius a. Both the line charge and the conducting cylinder are assumed to be infinitely long. Determine the position of the image charge.

Image: (1) the cylindrical surface at r=a an equipotential surface → the image must be a parallel line charge (ρi) inside the cylinder.(2) symmetry with respect to the line OP → the image must lie somewhere along OP. Say at point Pi, which is at a distance di from the axis.

4.4-2 Line charge and parallel conducting cylinder

Determine: ρi and di

Let us assume that ρi = - ρl, and electric potential at a distance r from a line charge of density ρl is

0 0

0

0 0

1ln

2 2

r r

rr r

rV E dr dr

r r

r0 : zero potential

At any point M on the cylindrical surface, the potential:

0 0

0 0 0

ln ln ln2 2 2

iM

i

r r rV

r r r

r0 for both ρi and ρl

Original charge Image charge

Triangles OMPi and OPM similar :

The image line charge ρi together with ρl, will make the dashed cylindrical surface equipotential.

Equipotential

2

i

ad

d

The second equipotential surface

Two wire transmission line

Electric field and equipotential lines of a transmission line pair

4.4-3 Point charge and conducting sphere

Q: A point charge Q is at a distance d from the center of a grounded conducting sphere of radius a (a<d). Determined the charge distribution induced on the sphere and the total charge induced on the sphere.

Image charge Qi can be equal to –Q?Triangles OMPi and OPM are similar.

The electric potential V at an arbitrary point

0 0

1 1( , ) ( ) ( )

4 4i iQ Q Q Q

aQQ Q adV R

R R R dR

The law of cosines 2 2 1/2[ 2 cos ]QR R d Rd 2 2

2 2 1/2[ ( ) 2 ( )cos ]iQ

a aR R R

d d

0 0

1 1( , ) ( ) ( )

4 4i iQ Q Q Q

aQQ Q adV R

R R R dR

The R-component of the electric field intensity ER

( , )( , )R

V RE R

R

2

2 2 3/2 2 2 2 2 3/20

cos [ ( / ) cos ]( , ) { }

4 ( 2 cos ) [ ( / ) 2 ( / ) cos ]R

Q R d a R a dE R

R d Rd d R a d R a d

(2) The total charge induced on the sphere is obtained by integrating ρs over the surface of the sphere.

22

.

0 0

Total-induced-charge sins s i

ads a d d Q Q

d

(1) To find the induced surface charge on the sphere, we set R=a

2 2

0 2 2 3/2

( )( , )

4 ( 2 cos )s R

Q d aE a

a a d ad

Induced surface charge is negative and that its magnitude is maxmium at θ=0 and minimum θ=π.

4.4-4 Charged sphere and grounded plane

4.5 Boundary-value problems in Cartesian Coordinates

Method of separation of variablesFor problems consisting of a system of conductors maintained at specified potentials and with no isolated free charges

Three types of boundary-value problems:(1)Dirichlet problems: the potential value V is specified everywhere on the boundaries;(2)Neumann problems: the normal derivative of the potential is specified everywhere on the boundaries;(3)Mixed boundary-value problems: the potential value is specified over some boundaries and the normal derivative of the potential is specified over the remaining ones;

2 0V Potential equation (free source):

Laplace’s equation for V in Cartesian coordinates

2 2 2

2 2 20

V V V

x y z

Apply method of separation of variables, V(x,y,z) can be expressed as:

( , , ) ( ) ( ) ( )V x y z X x Y y Z z

X(x),Y(y) and Z(z) are functions of only x, y and z, respectively.

2 2 2

2 2 2

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 0

d X x d Y y d Z zY y Z z X x Z z X x Y y

dx dy dz

2 2 2

2 2 2

1 ( ) 1 ( ) 1 ( )0

( ) ( ) ( )

d X x d Y y d Z z

X x dx Y y dy Z z dz

In order for Equation to be satisfied for all values of x, y, z, each of the three terms must be a constant.

kx2 is a constant to be determined from the boundary conditions

kx is imaginary, -kx2 is a positive real number

kx is real, -kx2 is a negative real number

2

2

1 ( )[ ] 0

( )

d d X x

dx X x dx

22

2

1 ( )

( ) x

d X xk

X x dx

22

2

( )( ) 0x

d X xk X x

dx

22

2

( )( ) 0y

d Y yk Y y

dy

22

2

( )( ) 0z

d Z zk Z z

dz

2 2 2 0 x y zk k k

Possible Solutions of

Q: Two grounded, semi-infinite, parallel-plane electrodes are separated by a distance b. A third electrode perpendicular to and insulated from both is maintained at a constant. Determine the potential distribution in the region enclosed by the electrodes.

With V independent of z , , ,V x y z V x y

In the x direction 00,V y V , 0V y

In the y direction ,0 0V x , 0V x b

2( ) kxX x D e

1( ) sinY y A ky

2 2 2 0 x y zk k k 2 2 2y xk k k k real number, so kx=jk

0zk 0Z z Bz-independence:

0 2 1, kx kxn nV x y B D A e sinky C e sinky

( , , ) ( ) ( ) ( )V x y z X x Y y Z z

, 0kxn nV x b C e sinkb

Should be satisfied for all values of x, only if

, 1, 2,3,...n

k nb

( , ) sinn

xb

n n

nV x y C e y

b

01 1

(0, ) (0, ) sinn nn n

nV y V y C y V

b

04,

0,n

Vn odd

C nn even

Solution of Electrostatic Problems

Method of imagesfree charges near conducting boundaries.

Method of separation of variablesfor problems consisting of a system of conductors maintained at specified potentials and with no isolated free charges.the solution can be expressed as a product of three one-dimensional functions, each depending separately on one coordinate variable only