Fall 2014 Notes 23 ECE 2317 Applied Electricity and Magnetism Prof. David R. Jackson ECE Dept. 1.

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

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

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