Electrostatic Boundary value

problems

Sandra Cruz-Pol, Ph. D.INEL 4151 ch6

Electromagnetics IECE UPRM

Mayagüez, PR

Last Chapters: we knew either V or charge distribution, to find E,D.

NOW: Only know values of V or Q at some places (boundaries).

Some applications

Microstrip lines capacitance Microstrip disk for microwave equipment

To find E, we will use:

Poisson’s equation:

Laplace’s equation: (if charge-free)

They can be derived from Gauss’s Law

02 V

vV 2

VE

ED v

Depending on the geometry:

We use appropriate coordinates:

cartesian:

cylindrical:

spherical:

vV 2

v

z

V

y

V

x

V

2

2

2

2

2

2

v

z

VVV

2

2

2

2

2

11

vV

r

V

rr

Vr

rr

2

2

2222

2 sin

1sin

sin

11

Procedure for solving eqs.

1. Choose Laplace (if no charge) or Poisson

2. Solve by Integration if one variable or by

3. Separation of variables if many variables

4. Apply B.C.

5. Find V, then E=-DV, D=eE, J=sE

6. Also, if necessary:

SnD S

dSJIdSQ S

P.E. 6.1 In a 1-dimensional device, the charge density is given by

If E=0 at x=0 and V=0 at x=a, find V and E.

Evaluating B.C.

vV 2

axov /

v

x

V

2

2

BAxa

xV o

6

3

xo

x aAa

xa

x

VE ˆ

2ˆ

2

0A

6

2aB o 33

3

6xa

a

xV o

xo a

a

xE ˆ

2

2

axov /

45o

P.E. 6.3 two conducting plates of size 1x5m are inclined at 45o to each other with a gap of width

4mm separating them as shown below. Find approximate charge per plate

if plates are kept at 50V potential difference and medium between them has permittivity of 1.5

Applying B.C. V(0)=0,

V(fo=45)=Vo=50

a

Va

VE

o

o ˆˆ1

01

2

2

22

V

V

02

2

V

o

oVV

ED o

ons

VD

)0(

)/ln(1

0

abLV

dzdV

dSQo

oL

z

b

ao

os

1m

BAV

P.E. 6.3 two conducting plates of size 1x5m are inclined at 45o to each other with a gap of width

4mm separating them as shown below.

permittivity of 1.5

Applying B.C. V(0)=0,

V(fo=45)=Vo=50)/ln( abL

V

QC

oo

)/ln( abLV

Qo

o

45o

a b

)/ln( abLV

QC

oo

mmmm

ao

226.52/45sin

2/4 nCVpFCVQ

F

mm

mmm

V

QC

o

o

o

22)50(444

10444

226.5

1000ln)5(

4/

5.1

12

detail

45o

a b

a

brecha

hipotenusa

opuestoo 2/

2

45sin

mmmm

ao

226.52/45sin

2/4

P.E. 6.5 Determine the potential function for the region inside the rectangular trough of

infinite length whose cross section is shown. The potential V

depends on x and y. Vo=100V, b=2a=2m,

find V and E at:a) (x, y)=(a, a/2)

b) (x, y)=(3a/2, a/4)

oVaybxV

ybxV

aybxV

ayxV

),0(

0)0,0(

0)0,(

0)0,0(

y

xb

aV=Vo

V=0

V=0V=0

02

2

2

2

y

V

x

V

P.E. 6.5 (cont.) Since it’s 2 variables, use Separation of Variables

y

xb

aV=Vo

V=0

V=0V=0

)()(),( yYxXyxV

0"" XYYX

Y

Y

X

X ""

0"

0"

YY

XX

einseparablVaYxXaxV

YYxXxV

bXyYbXybV

XyYXyV

o

)()(),(

0)0(0)0()()0,(

0)(0)()(),(

0)0(0)()0(),0(

Let’s examine 3 Possible Cases

.A l=0

.B l<0

C. l>0

Case A: If l=o

BAxX

0

0

A

B

0)()(

0)(

yYxXV

xX

y

xb

a

V=Vo

V=0

V=0V=0

0"

0"

YY

XX

This is a trivial solution, therefore l cannot be equal to zero.

0)(

0)0(

..

bX

X

CB

Case B: l<o

xBxBX

or

eAeAX xx

sinhcosh

:issolution general

21

21

0

0sinh

0

2

2

1

B

bB

B

2 0" 2 XX

This is another trivial solution, therefore l cannot be equal to zero.

0)(

0)0(

..

bX

X

CB

Case C: l>o

xgxgX

or

eCeCX

o

xjxjo

sincos

:issolution general

1

1

0sin

0

1

bg

go

2 0" 2 XX

0)(

0)0(

..

bX

X

CB

4,3,2,1

0sin

nb

n

b

22

sin)(

:solutions of seriesA

b

n

b

xngxX nn

Case C: l>o

yhyhY o sinhcosh

:issolution generalwith

10oh0)0(

..

Y

CB

0" 2 YY 4,3,2,1

nb

n

xhyYn sinh)( 1

b

yn

b

xnhgyYxXyxV nnnnn

sinhsin)()(),(

b

yn

b

xncyxV n

nn

sinhsin),(

1

By superposition, the combination is also a solution:

Cont.

b

an

b

xncVaxV n

non

sinhsin),(

1

dxb

xn

b

xm

b

ancdx

b

xmV

b

nn

o

b sinsinsinhsin

010

nm

nmdxmxnx

2/

0sinsin

0

B.C. at y=a

If we multiply by sin factor and integrate on x:

dxb

xm

b

ancdx

b

xnV

b

no

b 2

00

sinsinhsin

dxb

xn

b

anc

b

xn

n

bV

b

n

b

o

2cos1

2

1sinhcos

00

Orthogonality property of sine and cosine:

2

sinhcos1b

b

ancn

n

bVn

o

dxb

xn

b

anc

b

xn

n

bV

b

n

b

o

2cos1

2

1sinhcos

00

ban

n

byn

bxn

VyxV

n

o

sinh

sinhsin4),(

5,3,1

evenn

oddn

ban

n

V

c

o

n

0

sinh

4

Flux linesEquipotential lines

V=Vo

V=0

V=0

Find V(a,a/2) where Vo=100V, b=2a=2m

V

nn

ynxn

Vn

51.442

1sinh

2sinh

2sin400

5,3,1

Flux linesEquipotential lines

V=Vo

V=0

V=0

ban

n

ban

ban

VaaV

n

o

sinh

2sinhsin4

)2

,(5,3,1

Find E at (a,a/2)

yx

oddn

o ab

yn

b

xna

b

yn

b

xn

banb

VyxE ˆcoshsinˆsinhcos

sinh

14),(

yx ay

Va

x

VVE ˆˆ

V/mˆ25.99

ˆ...)0074.0035.01703.08192.9411.3127.1912.115(

ˆ2/

coshsinsinh

1400ˆ0

y

y

yoddn

x

a

a

ab

an

b

an

banb

aE

Resistance and Capacitance

Resistance

If the cross section of a conductor is not uniform we need to integrate:

Solve Laplace eq. to find VThen find E from its differentialAnd substitute in the above equation

S

l

SdE

ldE

I

VR

P.E. 6.8 find Resistance of disk of radius b and central hole of radius a.

S

oo

SdE

V

I

VR

011

2

2

2

2

2

z

VVV

01

V

BAV lnoVbV

aVBC

)(

0)(:

aab

VV o

ln/ln

ˆd

dVVE

ˆ

/ln ab

Vo

)/ln(

2

ab

tVSdEI o

S

ˆddzSd

t

ab

o2

)/ln(

a

t

b

Capacitance Is defined as the ratio of

the charge on one of the plates to the potential difference between the plates:

Assume Q and find V (Gauss or Coulomb)

Assume V and find Q (Laplace)

And substitute E in the equation.

FaradsldE

SdE

V

QC

l

S

Capacitance

1. Parallel plate

2. Coaxial

3. Spherical

Parallel plate Capacitor Charge Q and –Q

orx

s

xsn

aE

aD

ˆ

ˆ

Dielectric, e

Plate area, S

S

Qs

d

S

V

QC

S

Qddx

S

QldEV

dd

00

SESdEQ x

Coaxial Capacitor

Charge +Q & -Q

LESdEQ 2

abL

V

QC

ln

2

Dielectric, e

Plate area, S

S

Qd

S

QdSEV

dd

00

++

+

+

+

-

-

-

-

--

-

-

-

c

a

b

L

Qd

L

QldEV

a

b

ln2

ˆˆ2

Spherical Capacitor

Charge +Q & -Q

24 rEdSEQ r

baV

QC

114

ba

Qrdrr

r

QldEV

a

b

11

4ˆˆ

4 2

What is the Earth's charge?

The electrical resistivity of the atmosphere decreases with height to an altitude of about 48 kilometres (km), where the resistivity becomes more-or-less constant. This region is known as the electrosphere. There is about a 300 000 volt (V) potential difference between the Earth's surface and the electrosphere, which gives an average electric field strength of about 6 V/metre (m) throughout the atmosphere. Near the surface, the fine-weather electric field strength is about 100 V/m.

The Earth is electrically charged and acts as a spherical capacitor. The Earth has a net negative charge of about a million coulombs, while an equal and positive charge resides in the atmosphere.

Capacitors connection

Series

Parallel

21 CCC

21

111

CCC

Resistance

S

SdE

ldE

I

VR

ldE

SdE

V

QC S

Recall that:

Multiplying, we obtain the Relaxation Time:

Solving for R, we obtain it in terms of C:

RC

CR

So In summary we obtained:Capacitor C R= /e sC

Parallel Plate

Coaxial

Spherical

ba11

4

S

d

ab

L

ln

2d

S

Lab

2

ln

4

11

ba

P.E. 6.9 A coaxial cable contains an insulating material

of s1 in its upper half and another material with s2 in its lower half. Radius of central wire is a and of the sheath is b. Find the leakage resistance of length L.

s1

s2

Lab

RLab

R2

21

1

lnln

21

2121 ||

RR

RRRRR

21

1ln

La

b

R

They are connected in parallel

P.E. 6.10a Two concentric spherical capacitors with e1r=2.5 in its outer half and another material with e2r=3.5 in its inner half. The inner radius is a=1mm, b=3mm and c=2mm . Find their C.

e1

pFC 53.0

e2

c

We have two capacitors in series:

21

21

CC

CCC

ba

C

ba

C oror

114

114 2

21

1

P.E. 6.10b Two spherical capacitors with e1r=2.5 in its

upper half and another material with e2r=3.5 in its lower half. Inner radius is a=1mm and b=3mm. Find their C.e1

e2

2121 || CCCCC

ddErdSEQ sin2

We have two capacitors in parallel:

22 ErQ

ba

VQC oro 11

2/ 1

1

pF

ba

C 5.011

2 21

ba

Qrdrr

r

QldEV

a

b

11

2ˆˆ

2 2

Method of Images

Whenever the is a charge in the presence of a conductor. The conductor serves as a mirror.

Substitute the conductor for a plane at V=0 and the image.

The solution will be valid only for the region above the conductor.

Line charge above ground plane

22

11

ˆ2

ˆ2

o

L

o

LEEE

2222 )(

ˆ)(ˆ

)(

ˆ)(ˆ

2 hzx

zhzxx

hzx

zhzxxE

o

L

),0,(),,0(),,(

),0,(),,0(),,(

2

1

hzxhyzyx

hzxhyzyx

VVV

2

1ln2

o

Ldlo

L

o

L

2

21

1 2

ˆˆ

2

22

22

)(

)(ln

2 hzx

hzx

o

L