lecture 3 electrostatics and conductors

8/14/2019 Lecture 3 Electrostatics and Conductors

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1

Electrostatics

and

Conductors


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2Peter Gustav Lejeune Dirichlet(February 13th 1805 May 5th 1859)

Electrostatic propertiesof conductors

Carl Gottfried Neumann(May 7th 1832 March 27th 1925)


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3

Conductors in Equilibrium with Static Electric Fields

Property 1

Inside a conductor in electrostatic equilibrium, the electric

field is zero: E(x) = 0.Property 2

Inside a conductor in electrostatic equilibrium, the electricpotential is constant: V(x) = V0. The surface of the

conductor, which is also at potential V0, is anequipotential surface.Property 3

Inside a conductor in electrostatic equilibrium, the chargedensity is zero: (x) = 0 E(x) = 0. Any excess charge

on a conductor must therefore reside on its surface,described by a surface charge density .


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4

Conductors in Equilibrium with Static ElectricFields(cont)Property 4

At the surface of a conductor,

0tangential = tEE

0

normal

= nEE

There is an outward force per unit area on the surface ofa conductor, or electrostatic pressure, given by

( )

0

2

0

00 22

1limlim

=

=

A

AA

F

AA

The origin of this force is the Coulomb repulsion betweenlike charges:

0

22

0

0000

22

11lim

1limlim

=

=

lA

lAA

F

lAA


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5

Conductors in Equilibrium with Static ElectricFields(cont)Property 5 (Uniqueness theorem)

The potential V(x) throughout space is determined

(uniquely up to an additive constant) when(a) the external charge densities are specified, and

(b) either the potentials of all conductors are specified: V(x) = V0 Dirichlet boundary condition

The practical importance of the uniqueness theorem isthat when a potential function satisfying the equations

and boundary conditions has been found, by anymethod, then we need look no further any otherfunction satisfying these conditions can only differ by anadditive constant.

or the charge densities on all surfaces are specified:( ) ( ) 0 = xxn V

Neumann boundarycondition


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6

The Uniqueness Theorem for Laplaces Equation

Suppose V1(x) and V2(x) are both functions that satisfy

Laplaces equation in a volume V of space:

and that satisfy the same boundary conditions on theset of boundary surfaces S of the volume.

012 = V 02

2 = V

Consider the

function( ) ( ) ( )

2121

VVVV =xFand the integral

( ) ( ) ( ) 0 21213 === SSV dAVVVVdAxd nnnFxF

since both V1 and V2 satisfy the same boundary conditionson S:21 VV = Dirichlet boundary

condition

21 VV = nn

Neumann boundarycondition


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( ) ( ) ( )2121 VVVV = xF

The Uniqueness Theorem for Laplaces Equation(cont)Now, ( ) ( ) ( ) ( ) ( )21

2

212121 VVVVVVVV += xF

( )[ ] ( ) 033221 == VV xdxdVV xF

( ) ( )xx 2121 VVVV ==Dirichlet boundary condition

( ) ( )xExEnn 2121 == VVNeumann boundary condition

Therefore,

( ) 021 = VV

for all x.

( ) ( ) constant21 = xx VV

The integrand is never negative, so the only way theintegral can be 0 is if


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8

Property 6

The electric field is zero in

any empty cavity in aconductor in electrostaticequilibrium, regardless ofthe electric field outside theconductor.

Conductors in Equilibrium with Static ElectricFields(cont)

Property 7If a charged conductor has within it an empty cavity, thefield in the cavity is zero and the charge on the cavity wallis zero the net charge of the conductor lies entirely onthe outer surface.


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9

Electrostatic problems withrectangular symmetry


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10

Charged Plates

Consider two plane conducting plates 1 and 2 separatedby a vacuum.

>0

( ) 00 =V

( ) 0VdV =

Assume the plates each have a large areaA- translation invariance in directions parallelto the plates:

( ) ( )zVV =x

Laplacesequation ( ) 0

2 = xV

( ) 02

2=zV

dzd ( ) z

dVzV 0= kE 0

dVV ==

The surface charge densities

d

Vu

0001

== Ek ( ) d

Vl

0002

== Ek


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11

Charged Plates (cont)

The total charges on the conductors are equal andopposite:

00

1 V

d

AQ

= 002 Vd

AQ

=

Recall two conductors that carry equal but oppositecharges Q form a capacitor. The capacitance of thesystem

V

QC

where Vis the potential difference between theconductors.It follows from the above that

d

A

C 0=for a parallel plate capacitor.

Homework: Work through Example 1 @ Page 100


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12

Consider a capacitor witharbitrarily shapedconductors at potentials V

1

and V2 with respective

charges Q and +Q.

Charged Plates (cont)

Assuming there is noexternal field or othercharges anywhere, the total

electrostatic energy( ) VQQVQVVdAxVdU =+=== 2

1

2

1

2

1

2

121

3

where the difference of potential between the conductors

V V2 V1.


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13

External Point Charges. The Method of Images

Consider a point charge at distancez0 from a grounded

plane.

Forz< 0, V= 0 and E = 0; while forz 0, Vsatisfies thePoissons equation

( ) ( ) ( ) ( )00

2 1 zzyxqV

= x

with the boundarycondition ( ) 00,, =yxV

(0, 0,z0)


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External Point Charges. The Method of Images(cont)Method of image charge

( )( ) ( )

+++

++ =

20

2220

220

11

4 zzyxzzyx

qV x

( ) ( )( )

( )[ ]( )

( )[ ]

+++

+++

++

++

==232

0

22

0

232

0

22

0

0

4 zzyx

zzyx

zzyx

zzyxqV

kjikjixxE

On thesurfacez=0,

( )( )

kE

20,,

232

0

22

0

0

zyx

qzyx

++ =

It follows that the surface

charge density:

( )

( )232

0

22

0

2

0,,

zyx

qzyx

++

=

The total charge induced onthe plane:

( ) qdxdyyx =

0,,

When q is placed at (0, 0,z0

), an equal but opposite

charge is drawn onto the conductor from ground.


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15

Example 2

Consider a point-like electricdipole p located at the point(0, 0,z

0

) above the

grounded planez= 0.What is the asymptoticelectric field on thezaxisforz>>z0?Solution:Recall the potential of a point-like dipole:( ) 2

04

rV

=prx

It follows that the potential of the above configuration isgiven by

( ) ( ) ( )

+++

++

= 20222

0

220

0

4 zzyxzzyx

p

V

jrjr

x

( )( )

( )[ ]( )

( )[ ]

+++

++

=232

0

22232

0

220

033

4 zzyxzzyx

p jrjrjrjrxEand


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16

Example 2 (cont)

Solution (cont):

For points on thezaxis,

( )( ) ( )

jE 11

4,0,0

3

0

3

00

0

+

+

=zzzz

pz

( ) jjE 2

3114

,0,04

0

00

3

0

3

0

3

0

0

z

zp

z

z

z

z

z

pz

++

=

Forz>>z0,

the far field is that of a quadrupole.

l i l


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Consider a point charge +q located at the point (0,y0,z0),

with mutually perpendicular planesy= 0 andz= 0 both

grounded.

Multiple Images

Recall the potential of a point charge:

( )( ) ( )

2

0

2

0

20

00

1

4,,0

zzyyx

qzyV

++ =

M l i l I ( )


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18

Multiple Images (cont)

( )

( ) ( )2

0

2

0

2

0

1

4 zzyyx

qV

++

=x

It follows that the potential of the above configuration isgiven by

( ) ( ) 202

0

20

1

4 zzyyx

q

+++

( ) ( )2

0

2

0

20

14 zzyyxq

+++

( ) ( )2

0

2

0

20

1

4 zzyyx

q

++++ +



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Ch d S h


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20

Charged Spheres

Consider a conducting sphere of radius a that carries acharge Q0.The potential inside the conductor, i.e., for r< a, V= V0.

For r a, Vsatisfies the Laplaces equation( ) 02 = xV 0

1 22

=

dr

dVr

dr

d

r

with boundary conditions

( ) 0VaV = ( ) 0=V

It follows that

02

2

2

=+dr

dV

rdr

Vd

( ) B

r

ArV += withA = aV0 and B = 0.

( ) ( ) rrxxE 2

00

r

aV

r

aV

dr

dV =

==

( ) 20

002

0

4 a

aV

aa

Q

=== Er aQ

V 0

0

0 4 =

Ch d S h ( t)


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21


( ) 0VaV = ( ) 0=bV

It follows that

02

2

2

=+dr

dV

rdr

Vd

( ) Br

ArV += withA and B satisfying

Charged Spheres (cont)

Consider two concentricconducting spheres.

For a r b, Vsatisfies theLaplaces equation

( ) 0VBa

AaV =+= ( ) 0=+= B

b

AbV

or ( )( ) ( )

( )rab

rbaV

br

rbA

b

A

r

ArV

=== 0

Ch d S h ( t)


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22

Charged Spheres (cont)

( ) ( )( )

( )rxxE 0

==rab

rbaV

dr

dV

( )( )

rE 2

0

rababVr

=

The surface chargedensities

( ) ( )aab

bV

aa ==0

00 Er

( ) ( )( )bab

aVbb

== 000 Er


002 44 Vab

abaQ aa

== 0

02 44 Vab

abbQ bb

==

The capacitance of the two concentric

spherical conductors ab

ab

V

Q

C

=

04

S h i l P bl ith D d


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23

Spherical Problems with Dependence on Laplaces equation ( ) 0

2 = xV

reduces to

( ) ( ) 0,sinsin1,1

2

2

2 =

+

rV

rrV

rr

rr

0sin

cos122

2

22

2

=

+

+

+ VV

rr

V

rr

Vor

It can be shown that the following is a solution:

( ) +++= coscos

,2

Drr

CB

r

ArV

The final term corresponds to a field uniform throughoutspace:

( ) ( ) ( )r cos1

coscos

= Dr

rDr

rDr

( ) kr

sincos DD ==

E ample 4


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Example 4

( ) kxE 0appl E=

Consider a groundedconducting sphere, of radius aand centered at the origin, inan externally applied field inthezdirection:

The presence of the spherechanges the field. What arethe potential V(r, ), fieldE(x), and surface chargedensity ( )?Solution:

( ) +++= coscos

,2

Drr

CB

r

ArV

0,0 == BA 0ED = 03EaC=

Example 4 (cont)


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25

Solution: (cont)

( ) = coscos, 020

3

rEr

EarV

( )

+=

=cos1

2,

03

3

Er

a

r

VrE

r

For r< a, V= 0; while for r

a,

It follows that

( )

=

= sin1

1, 03

3

Er

aV

rrE

Example 4 (cont)

( ) ( ) == cos3 000 EaEr

The total surface charge

( ) ( ) 0sin20tot ==

adaQ

External Charges


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External Charges

Consider a grounded spherical conductor of radius R,centered at the origin; and an external point charge qlocated at (0, 0,z

0

):

( )

+

++ = cos2cos241

, 02

02

020

20 zrzr

q

rzzr

q

rV

For r R

External Charges (cont)


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27


AtA, ( ) 0224

10,

0

2

0

2

0

2

0

20

=

+

++

=zRzR

q

RzzR

qRV

( ) 0224

1,

0

2

0

2

0

2

0

20

=

++

+++

=zRzR

q

RzzR

qRV

( ) ( )RzqzRq = 00

At B,

( ) ( )RzqzRq +=+ 00

It follows that

200

0

0

0

0 RzzRzRz

zRzR =+=+

( ) qz

RqRzq

z

RRq

0

0

0

2

=+=

+

( )

+

+ =

cos2cos2

1

4,

0

2

0

2

00

2

0

20 zrzrz

R

rzzr

qrVAnd

++ = cos2cos21

4 20

42200

20

20 rRzRrz

R

rzzr

q

conjugate

points



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( )

+

+

=

2

1

0

2

22

0

4

0

2

1

0

2

2

0

0

cos2

1cos2

11

4,

rz

R

rz

R

rz

R

r

z

r

z

r

qrV

The first two leading terms in the potential for the far field

( )

+

+

cos1cos1

1

4,

0

2

0

0

0 rz

R

rz

R

r

z

r

qrV

+

= cos1

4

11

4

103

0

3

2

000

qzz

R

rq

z

R

r

( )( )

( )( )

+

+

=

32

0

422

0

20

20

3

0

2

0

2

0

0 cos2

cos

cos2

cos4

,rRzRrz

RzrzR

rzzr

zrqrEr

Electric field

( )

( ) ( )

+

+

=

320

4220

0

3

30

20

2

0

0 cos2

sin

cos2

sin

4,

rRzRrz

zR

rzzr

zqrE

External Charges (cont) Pg 115


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29

External Charges (cont )

( )( )

( )

+

+

=3

3

0

422

0

20

20

3

0

2

0

2

0

cos2

cos

cos2

cos

4 RzRRz

RzRzR

RzzR

zRq

The surfacecharge density:

( ) ( )= ,0 REr

( )

( )

( )

+

+

= 3022

0

0

2

0

30

20

2

0

2

cos2

cos

cos2

cos

4 RzRz

Rzz

RzzR

RzR

R

q

( )3

0

2

0

2

2

0

2

cos24 +

=RzzR

zR

R

q

The total charge onthe conductor:

( ) ( ) qz

RRdRQ

00

tot sin2 ==

( )

( )

( )1

10

2

0

20

2

0

21

1 2

3

020

2

2

0

2

2

1

222 +

=

+

=

uRzzRzzRq

uRzzR

duzRqR

Pg 115 116


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Electrostatic problems withcylindrical symmetry

Charged Lines and Cylinders


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31

Charged Lines and Cylinders

Consider a charged line with constant linear chargedensity .There are two symmetries: translation invariance inz, the

coordinate along the line; and rotational invariance in ,the azimuthal angle.

Laplacesequation

( ) 02 = xV 01

=

dr

dVr

dr

d

r0

12

2

=+dr

dV

rdr

Vd

which has thegeneralsolution:

( ) Br

rArV +

=

0

ln

( ) ( ) rrxxE r

A

dr

dV

V ===

( )00 2

12

=

== Alrlr

AdAxE

Charged Lines and Cylinders (cont)


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32

Charged Lines and Cylinders (cont )

Consider a conducting cylinder of radius a which carriesthe constant line charge density spread uniformly onits surface:

Laplacesequation

( ) 02 = xV 01

2

2

=+dr

dV

rdr

Vd

which has thegeneral

solution:

( ) Ba

rArV +

= ln

( ) ( ) rrxxE r

A

dr

dVV ===

( )00 2

12

=

==

Alrl

r

AdAxE

aall == 22

In the region r a, Vsatisfies



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Consider two concentric conducting cylinders:


In the region a r b, Vsatisfies

Laplaces equation0

12

2

=+dr

dV

rdr

Vd


( ) 0VaV = ( ) 0=bV

It follows that ( ) Bb

rArV +

= ln withA and B

satisfying( ) 0ln VB

b

aAaV =+

= ( ) 0== BbV

or ( )( )

( )0

ln

lnV

ab

rbrV =



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( )( )

rE 1ln

0

rabVr =

( ) ( )( )

( )rxxE

ln

ln0

== V

ab

rb

dr

dV

The surface chargedensities

( ) ( )aab

V

aa ln0

00 == Er

( ) ( )( )bab

Vbb

ln 000 == Er


( ) 00

ln

22 V

ab

lalQ aa

==

( ) 00

ln

22 V

ab

lblQ bb

==

The capacitance per unit length of the

two concentric cylindrical conductors ( )abVl

Q

C ln

2 0

=

Two Infinitely Long Line Charges


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Consider two infinitelylong line charges withequal but opposite linearcharge densities + and , parallel to thezaxis,a distance 2dapart.

Two Infinitely Long Line Charges

The potential at (r, ):

( ) + +

= rrrV ln

2ln

2,

00

=+

r

rln

2 0

( )+++

=

cos2

cos2ln

4,

22

22

0 drdr

drdrrV

( )

( )

( ) 22

22

0222

222

0ln42

2

ln4, ydx

ydx

dxdyx

dxdyx

yxV +

++

=++

+++

=

Two Infinitely Long Line Charges (cont)


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Two Infinitely Long Line Charges (cont )

( )( )

( ) 22

22

0

222

222

0

ln42

2ln

4,

ydx

ydx

dxdyx

dxdyxyxV

+++

=+++++

=

The equation for an equipotential curve:( )

( )022

22

0

ln4

Vydx

ydx=

+++

( )

( )

=+++ 00

22

22 4exp

V

ydx

ydx

( )[ ] ( )[ ]22002200 2

exp2

exp ydxV

ydxV

++

=+

( ) 02cosh22sinh 0022200 =

++

dxV

ydxV

02csch2coth 200222

00 =+

yVdVdx

( ) 2220 Ryxx =+


Two Infinitely Long Line Charges (cont)


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Two Infinitely Long Line Charges (cont )

The potential of a line dipole

( )+++

=

cos2cos2ln

4,

22

22

0 drdrdrdrrV

+

++

=

2

2

2

2

0

cos21ln

cos21ln

4 r

d

r

d

r

d

r

d

++

2

2

2

2

0

cos2cos2

4 r

d

r

d

r

d

r

d

( ) rr

d

rV 00 2

2

cos2

,

=

rp

ip 2 d=In the dipole limit, where d 0 and with2 d remaining finite, we have a line dipoledescribed by the vector dipole moment perunit length:

( )+++

=cos2

cos2ln

4,

22

22

0 drdr

drdrrV

Cylindrical Problems with Dependence on


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Cylindrical Problems with Dependence on Laplaces equation ( ) 0

2 = xV

reduces to

( ) ( ) 0,1,1 2

2

2 =+

rV

rrV

rr

rr

011

2

2

22

2

=

++

V

rr

V

rr

Vor

It can be shown that the following is a solution:

( ) +++= coscos

ln, Drr

CBrArV

The final term corresponds to a field uniform throughoutspace:

( ) ( ) ( )

= cos

1coscos Dr

rDr

rDr r

( ) ir

sincos DD ==

Example 5


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39

( ) ixE 0appl E=

Consider a groundedconducting cylinder, of radius aand centered at the origin, in

an externally applied field inthexdirection:

The presence of the spherechanges the field. What arethe potential V(r, ), fieldE(x), and surface chargedensity ( )?

Example 5

Solution:

( ) +++= coscosln, Drr

CBrArV

0,0 == BA 0ED = 02EaC=

Example 5 (cont) St. Elmos


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40

Example 5 (cont )

Solution: (cont)

( ) = coscos, 002

rEr

EarV

For r< a, V= 0; while for r a,

It follows that

( )

+=

= cos1,

02

2

Er

a

r

VrE

r

( )

=

= sin1

1, 02

2

Er

aV

rrE

( ) ( ) == cos2 000 EaEr

The total surface charge

( ) 0

2

0tot ==

ad

St. Elmo sfire

External Line Charge


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41

External Line Charge

Considery= 0 to be a grounded conducting plane. Thereis an infinitely long line charge above it. The line chargeis parallel to thezaxis, intersects thexyplane at the

point (0,y0) and has constant linear charge density + .

( )( ) ( )

0

2

0

2

00

2

0

2

0

ln2

ln2

,r

yyx

r

yyxyxV

++

++

=

( ) ( )( )

( )( )

++++

++

=

2

0

2

0

2

0

2

0

0

2

,yyxyyx

yyxyyxyx jijiE

( )2

0

2

0

0

0,

yx

yx

+ =

jE ( ) ( )

2

0

2

00

0,,yx

yxzx

+== jE

The induced charge per unit length on the plane

( ) =+

=

20

2

0,yx

dxydxzx

External Line Charge (cont)


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42

External Line Charge (cont )

Consider an infinite linecharge + in the presenceof an infinite isolated

charged conducting cylinderwith linear charge density .

( ) ( + = cos2ln21

, 02

0

2

0rxxrrV

( ) ( ) ( )[ ]000

lnln210, xRRxRV +

=

( ) ( )[ ] ( )=+++

= ,lnln2

100

0

RVxRxR

It follows if = , that 2

00 Rxx =

)++ cos2ln 0202 xrxrconjugate

points

External Line Charge (cont)


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43

External Line Charge (cont )

( )

=

00

ln2

,x

RRV

( )++

=

cos2

cos2ln

4,

0

2

0

2

0

2

0

2

0 rxxr

xrxrrV

( )++

= cos2cos2

ln4 020

220

0

242

0

2

0 rxxrx

rxRRxr

( )( )+

+ =

cos2

cos2ln

4 02

0

22

0

0

22

0

222

0 rxxrx

rxRxrRR


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lecture 3 electrostatics and conductors

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