Electrostatics 2

UCF Boundary Conditions

tt EE 12

. gives This .0 as 0 volume

since 0 charge, volumefor or ;0 charge, noFor

12

volumeinside

volumeinside

nn DDhQQ

021 LELE tt0 dLE

20h

0Volume

QdSD2

0)( 12n EEa 1LMedia 1

Media 2

E1

E2

E1t

E2t

0h1

S

Media 1

Media 2

D1

D2

D1n

D2n volumeinside12 QSDSD nn

SQ S volumeinside charge, surfaceFor SSnn DD )(or 1212 DDan

(1)

(2)

na

na

UCF Boundary Conditions for Potential

2V21 VV 2

11111

222

VV

EDED 22 1Media 1

Media 2

S

S

nV

nV

nV

nV

22

11

11

22 )()(

. charge, surface no is thereIf 2211 nV

nV

S )( 12 DDan

(1)

(2)1V

Potential should be continuous across the boundary

na

UCF Perfect Electric Conductor (PEC)

0Ean 0tE

0E

Media D

nV

S Dan

SDan

(1)

(2)

na

PEC

or

0BAB A B AV V V E dLOn PEC body (including boundary),

PEC is equal potential.

UCF Capacitance

+Q

-Q

+

-V

VQC

UCF Energy Stored in Electrostatic Field

ED 21

ew

V ee dVwW

Energy density

Total energy

UCFPoissons and Laplaces Equations

02

2

V

V V

0 V

VV D

From electrostatic equations:

VDE

0

In simple media

V E

Or

V ED

For source free region, we have

VV Then

If the dielectric distribution is uniform

Laplaces Equation

Poissons Equation

UCFUniqueness Theorem

02

2

V

V V

na

VV D

For

nV

nV

If on the boundary, V or is given, the solution is unique.

Laplaces Equation

Poissons Equation

or

given

V given

UCFProof of Uniqueness Theorem (1)

)( )()()()()(

have we,let weif

formulamath From

0or 0

have wegiven, is or either since boundary, On the

0)( have we

Equations, sLaplace'or sPoisson'satisfy and solutions twoAssuming

221

22121

2212121

21

22

boundary on 21

boundary on 21

212

21

VV

VVVVVVVVVV

VV

nV

nVVV

nVV

VV

VV

Proof:

UCFProof of Uniqueness Theorem (2)

solutions. identical twogiving Therefore, given. is when

0boundary on the since zero be toevaluatedeasily becan constant Theconstant

or0)(

have Then we

condition)boundary (from 0 ))((

Theorem) sGauss' (from )()(

)()()(

have wee,over volumequation above thegIntegratin

21

21

21

21

S

2121

S2121

21212

21

VVV

VVVV

VV

dSnV

nVVV

VVVV

dVVVVVdVVV

dS

UCF Laplaces Equations

Rectangular (Cartesian) Coordinate

Polar (Cylindrical) Coordinate

Spherical Coordinate

UCF Example 1 (1)Find the potential distribution between two parallel plates. Assume the two plates are large enough so that

0 and 0

yV

xV

z2

z1

z

d

V1

V2

UCF Example 1 (2)

02 VFrom Laplaces equation

0 and 0

yV

xV

022

dzVd

022

2

2

2

2

zV

yV

xV

)( and )( 2211 VzVVzV

Solution:we have

in Cartesian coordinate.

Assuming the two plates are large enough so that

we have

The boundary conditions are:

(1)

(2)

UCF Example 1 (3)

zdVzV 2)( 12

1221

12

12)(zzzVzVz

zzVVzV

The general solution for (1) is

have we,0 and 0 If 11 Vz

22

11

VBAzVBAz

12

1221

12

12 and zzzVzVB

zzVVA

BAzzV )( (3)Inserting (2) into (3) results in

from which, we have

Therefore

linear distribution

This page can be done using MatLab dsolve.

UCF Example 1 (4)

zaED dV2

zaE dVV 2

Find capacitance between two parallel plates.

zdVzV 2)(

0 and 0

yV

xVz2=d

0

z

d0

V2

Then

)()(

2

2 top

22 top

dS

VQC

SdVSQ

dV

dV

S

S

zz aa

zaS

S

UCF Example 2

0 ,0

r

02

1 0

0sin

V)V()V(

ddV

dd

0cossin 22

ddV

dVd

To use dsolve, we need to change the equation to: a