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

VEED v

Depending on the geometry:

We use appropriate coordinates:cartesian:

cylindrical:

spherical:

vV 2

v

zV

yV

xV

2

2

2

2

2

2

v

zVVV

2

2

2

2

2

11

vV

rV

rrVr

rr

2

2

2222

2 sin1sin

sin11

Procedure for solving eqs.

1. Choose Laplace (if no charge) or Poisson 2. Solve by Integration if one variable or by3. Separation of variables if many variables4. Apply B.C.5. Find V, then E=-DV, D=E, J=sE6. 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

xV

2

2

BAxaxV o

6

3

xo

x aAaxa

xVE ˆ

2ˆ

2

0A

6

2aB o 333

6xa

axV o

xo a

axE 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(o=45)=Vo=50

aVaVE

o

o ˆˆ1

012

2

22

VV

02

2

V

o

oVV

ED o

ons

VD )0(

)/ln(1

0

abLVdzdVdSQo

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 1.5Applying B.C. V(0)=0,V(o=45)=Vo=50

)/ln( abLVQ

Coo

)/ln( abLVQo

o

45o

a b

)/ln( abLVQ

Coo

mmmma o 226.52/45sin

2/4 nCVpFCVQ

F

mmmmm

VQ

C

o

o

o

22)50(44410444

226.51000ln)5(

4/5.1

12

detail

45o

a b

abrecha

hipotenusaopuestoo 2/

245sin

mmmma o 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)

oVaybxVybxV

aybxVayxV

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

y

xb

aV=Vo

V=0

V=0V=0

02

2

2

2

yV

xV

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

YY

XX ""

0"0"

YYXX

einseparablVaYxXaxVYYxXxV

bXyYbXybVXyYXyV

o

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

Let’s examine 3 Possible Cases

A. =0B. <0C. >0

Case A: If =o

BAxX

00

AB

0)()(0)(

yYxXVxX

y

xb

a

V=Vo

V=0

V=0V=0

0"0"

YYXX

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

0)(0)0(

..

bXX

CB

Case B: <o

xBxBXor

eAeAX xx

sinhcosh

:issolution general

21

21

00sinh

0

2

2

1

BbB

B

2 0" 2 XX

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

0)(0)0(

..

bXX

CB

Case C: >o

xgxgXor

eCeCX

o

xjxjo

sincos

:issolution general

1

1

0sin0

1

bggo

2 0" 2 XX

0)(0)0(

..

bXX

CB

4,3,2,1

0sin

nb

nb

22

sin)(

:solutions of seriesA

bn

bxngxX nn

Case C: >o

yhyhY o sinhcosh:issolution generalwith

1 0oh0)0(..Y

CB

0" 2 YY 4,3,2,1

nb

n

xhyYn sinh)( 1

byn

bxnhgyYxXyxV nnnnn

sinhsin)()(),(

byn

bxncyxV n

nn

sinhsin),(1

By superposition, the combination is also a solution:

Cont.

ban

bxncVaxV n

non

sinhsin),(1

dxb

xnb

xmb

ancdxb

xmVb

nn

o

b sinsinsinhsin010

nmnm

dxmxnx2/

0sinsin

0

B.C. at y=a

If we multiply by sin factor and integrate on x:

dxb

xmb

ancdxb

xnVb

no

b 2

00

sinsinhsin

dxb

xnb

ancb

xnnbV

b

n

b

o

2cos121sinhcos

00

Orthogonality property of sine and cosine:

2

sinhcos1 bb

ancnn

bVn

o

dxb

xnb

ancb

xnnbV

b

n

b

o

2cos121sinhcos

00

bann

byn

bxn

VyxVn

o

sinh

sinhsin4),(5,3,1

evennoddn

bann

V

co

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

bann

ban

ban

VaaVn

o

sinh

2sinhsin4)

2,(

5,3,1

Find E at (a,a/2)

yx

oddn

o ab

ynb

xnab

ynb

xn

banb

VyxE ˆcoshsinˆsinhcossinh

14),(

yx ayVa

xVVE ˆˆ

V/mˆ25.99

ˆ...)0074.0035.01703.08192.9411.3127.1912.115(

ˆ2/coshsinsinh

1400ˆ0

y

y

yoddn

x

a

a

aban

ban

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

IVR

s

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

S

oo

SdEV

IVR

s

0112

2

2

2

2

zVVV

01

V

BAV lnoVbV

aVBC

)(0)(

:

aabVV o ln

/ln

ˆddVVE

ˆ

/ln abVo

)/ln(2

abtVSdEI o

S

ss

ˆddzSd

s tab

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

VQC

l

S

Capacitance

1. Parallel plate2. Coaxial 3. Spherical

Parallel plate Capacitor Charge Q and –Q

orx

s

xsn

aE

aD

ˆ

ˆ

Dielectric,

Plate area, S

SQ

s

dS

VQC

SQddx

SQldEV

dd

00

SESdEQ x

Coaxial Capacitor Charge +Q & -Q

LESdEQ 2

abL

VQC

ln

2

Dielectric,

Plate area, S

SQd

SQdSEV

dd

00

++

+

+

+

--

-

--

-

-

-

-

c

ab

LQd

LQldEV

a

b

ln2

ˆˆ2

Spherical Capacitor Charge +Q & -Q

24 rEdSEQ r

baVQC

114

ba

Qrdrrr

QldEVa

b

114

ˆˆ4 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 connectionSeries

Parallel

21 CCC

21

111CCC

Resistance

S

SdE

ldE

IVR

s

ldE

SdE

VQC S

Recall that:

Multiplying, we obtain the Relaxation Time:

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

s

RC

CR

s

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

Parallel Plate

Coaxial

Spherical

ba11

4

Sds

ab

Lln

2dS

Lab

s2

ln

s4

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

ss

21

2121 ||

RRRRRRR

21

1ln

ss

Lab

R

They are connected in parallel

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

1

pFC 53.0

2

c

We have two capacitors in series:

21

21

CCCCC

ba

C

ba

C oror

114

114 2

21

1

P.E. 6.10b Two spherical capacitors with 1r=2.5 in its

upper half and another material with 2r=3.5 in its lower half. Inner radius is a=1mm and b=3mm. Find their C.1

2

2121 || CCCCC

ddErdSEQ sin2

We have two capacitors in parallel:

22 ErQ

ba

VQC oro 11

2/ 11

pF

ba

C 5.0112 21

ba

Qrdrrr

QldEVa

b

112

ˆˆ2 2

Method of ImagesWhenever 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 plane2

21

1

ˆ2

ˆ2

o

L

o

LEEE

2222 )(ˆ)(ˆ

)(ˆ)(ˆ

2 hzxzhzxx

hzxzhzxxE

o

L

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

2

1

hzxhyzyxhzxhyzyx

VVV

2

1ln2

o

Ldlo

L

o

L

2

21

1 2ˆˆ

2

22

22

)()(ln

2 hzxhzx

o

L