Fall 2014

Notes 23

ECE 2317 Applied Electricity and Magnetism

Prof. David R. JacksonECE Dept.

1

Boundary Value Problem

2 , , 0x y z

, ,B x y z

, ,x y z is unique.

Uniqueness theorem:

On boundary:

Goal: Solve for the potential

function inside of a region, given the value of the potential function

on the boundary.

(Please see the textbooks for a proof of the uniqueness theorem.)

(no charges)

2

As long as our solution satisfies the Laplace equation and the

B.C.s, it must be correct!

Example: Faraday Cage Effect

0v

Guess:

B = V0 = constant

Check:

Hollow PEC shell0

Prove that E = 0 inside a hollow PEC shell (Faraday cage effect).

0, ,x y z V r V

2 20

0

0 in

on S

V V

V

Therefore: The correct solution is

V

0, ,x y z V

S

3

Note: We can make any guess

that we wish, as long as our final solution satisfies

Laplace’s equation and the boundary conditions.

Example (cont.)

0v

Hollow PEC shell

0 V

0, ,x y z V

S

Hence E = 0 everywhere inside the hollow cavity.

0, , 0E x y z V

B = V0 = constant

4

(Faraday cage effect)

Example

Solve for (x, y, z)

Assume: , ,x y z x

+ - r h

xV0

0V

0

R

5

Ideal parallel-plate capacitor

Note: We can make any assumptions that we wish, as long as our final solution satisfies the boundary conditions.

Example (cont.)

2 2 2

2 2 20

x y z

Hence

2 0

Solution:

1

1 2

x C

x C x C

2

20

x

x

6

Example (cont.)

0 1 2 0 1 0 /h V C h C V C V h

1 2 20 0 0 0 0C C C

Hence we have

01

2 0

VC

hC

0 V, ,V

x y z xh

The solution is then

7

+- r h

xV0

0V

0

R

1 2x C x C

0 :x

:x h

Example (cont.)Calculate the electric field:

0ˆ ˆx V

E x xx h

From previous notes: 0

0

B

A

x

h

x

V E dr

E dx

E h

0x

VE

hso

8

0 V/mˆV

E xh

0, ,V

x y z xh

0ˆ V/mV

E xh

Hence

Example

2 2

2 2 2

1 10

z

, , z Assume

9

+ -

V0

0V

0

R

2 0 0

rWedge

Insulating gap

Example (cont.)

2

20

0 : 1 2

2

0 0

0

C C

C

0 : 1 0 2 0

01

0

C C V

VC

1 2C C Hence

10

Example (cont.)

0

0

, , VV

z

1ˆˆ ˆ

1

E

zz

0

0

1ˆ V/mV

E

Hence

Find the electric field:

We then have

11

Example (cont.)

0

0

1ˆ V/mV

E

12

Flux plot

+ -

V0

0V

0

r

Insulating gap

Example

1 1 2

2 1 2

c cx

d x d

0 :x

:x h 0 1 2 0h V d h d V

(Four unknowns)

+ - r1 h1

h2 xr2V0

0V

0

h

13

Two-layer capacitor

20 0 0c

Example (cont.)

1 1

2 1 0 1 1 0

c x

d x V d h d x h V

Hence:

1 :x h

1 1 1 1 0c h d h h V

so

We need two more equations: use interface boundary conditions.

BC #1

The potential is continuous across the boundary.

2

1

1 2 0r

r

E dr

21

Now there are two unknowns (c1 and d1).

(The path length is zero!)

14

1 1 1 2 0c h d h V

Example (cont.)BC #2:

1 2x xD D

To calculate Ex , use E

1 1 2 1r rc d Hence we have

1 :x h

xEx

1 2

1 2r rx x

Therefore

15

1 1 2 2r x r xE E

1 1

2 1 0

c x

d x h V

The normal component of flux density is continuous.

Example (cont.)

Therefore

11 1 2 0

2

r

r

c h h V

0 01 1

1 21 2 1 2

2 1

,r r

r r

V Vc d

h h h h

or

Hence we have

Also,

16

11 1 1 2 0

2

r

r

c h c h V

11 1

2

r

r

d c

1 1 1 2 0c h d h V

1 1 2 1r rc d

1 1

2 1 0

c x

d x h V

Example (cont.)

01 1

11 2

2

00 12

21 2

1

, 0

,

r

r

r

r

Vx x h

h h

Vh V h x hx

h h

0 0 21 1 1

1 2 2 111 2

2

0 0 12 12

1 2 2 121 2

1

0 0 1 21 2

1 2 2 1

ˆ ˆ, 0

ˆ ˆ ,

ˆ, 0

r

r rr

r

r

r rr

r

r r

r r

V VE x x x h

h hh h

V VE x x h x h

h hh h

VD D x x h

h h

We now find the electric fields and flux density:

17

Example (cont.)

0 0 1 21 0

1 2 2 1

0 0 1 22

1 2 2 1

ˆ

ˆ

bot r rs x

r r

top r rs x h

r r

Vx D

h h

Vx D

h h

We now find the surface charge densities on the plates.

ˆs D n Use

18

+ - r1 h1

h2 xr2V0

0V

0

h

++++++++++++++++

- - - - - - - - - - - - - - - -