Chapter 2 Electrostatics

2.0 New Notations

2.1 The Electrostatic Field

2.2 Divergence and Curl of Electrostatic Field

2.3 Electric Potential

2.4 Work and Energy in Electrostatics

2.5 Conductors

2.0 New Notations

R

2.1 The Electrostatic Field

2.1.1 Introduction

2.1.2 Coulomb’s Law

2.1.3 The Electric Field

2.1.4 Continuous Charge Distribution

2.1.1 Introduction

The fundamental problem for electromagnetic to solve is to calculate the interaction of charges in a given configuration. That is, what force do they exert on another charge Q ? The simplest case is that the source charges are stationary.

Principle of Superposition: The interaction between any two charges is completely unaffected by the presence of other charges.

1 2 3F F F F

R

is the force on Q due to iqiF

2.1.2 Coulomb’s Law

the permittivity of free space

The force on a charge Q due to a single point charge q is given by Coulomb`s law

20

212

0 2

1 ˆ ˆ 4

8.85 10

Q qqQ

F R R r r R RR

c

N m

2.1.3 The Electric Field

1 2

1 21 22 20 1 2

31 21 2 32 2 20 1 2 3

1 ˆ ˆ 4

ˆ ˆ ˆ 4

F F F

q Q q QR R

R R

qq qQR R R

R R R

F QE

20 1 1

1 ˆ( )4

Ni

ii

qE P R

R

the electric field of the source charge

2.1.4 Continuous Charge Distributions

( )1

i

nq

i

~ ~~ ( ) dlline

( ) dasurface

( ) dvolume

20 1

1 ˆ4

Ni

ii i

qE P R

R

20

ˆ14

line

RE P dR

20

ˆ14

surface

RE P daR

20

ˆ14

volume

RE P dR

12 2 2 2

32 2 20 2

ˆˆ ˆ( , , )

ˆˆ ˆ

ˆ

1, ,4 [ ]volume

x x i y y j z z kx y z dx dy dz

x x y y z z

R x x i y y j z z k

RR x x y y z z R

R

E x y z

2.1.4 (2)

20

ˆ1( , , ) ( , , )4 volume

RE x y z x y z dx dy dzR

, ,p x y z , ,x y z the source point is atthe test point is atExample:

20

ˆ1 ( , , )4 volume

R R x y z dx dy dzRR

30

1 ( , , )4 volume

R x y z dx dy dzR

2.1.4 (3)

Example 2.1 Find the electric field a distance z above the midpoint of a straight of length 2L, which carries a uniform line charge

20

1ˆ2 ( )cos

4

dxdE z

R

2 2 3 200

2 2 200

2 20

1 2

4 ( )

2

4

1 2

4

L

L

zE dx

z x

z x

z z x

L

z z L

Solution:

(1) z>>L2

0

1 2

4

LE

z

(2) L

0

1 2

4E

z

R

2.2 Divergence and Curl of Electrostatic Fields

2.2.1 Field Lines and Gauss’s Law

2.2.2 The Divergence of E

2.2.3 Application of Gauss’s Law

2.2.4 The Curl of E

A single point charge q, situated at the origin

2.2.1 Fields lines and Gauss’s law

20

1ˆ( )

4

qE r r

r

Because the field falls off like ,the vectors get shorter as I go father away from the origin,and they always point radially outward. This vectors can be connect up the arrows to form the field lines. The magnitude of the field is indicated by the density of the lines.

21 r

strength length of arrow strength density of field line

1.Field lines emanate from a point charge symmetrically in all directions.2.Field lines originate on positive charges and terminate on negative ones.3.They cannot simply stop in midair, though they may extend out to infinity.4.Field lines can never cross.

2.2.1 (2)

Since in this model the fields strength is proportional to the number of lines per unit area, the flux of ( ) is proportional to the the number of field lines passing through any surface . The flux of E through a sphere of radius r is:

22

0 0

1 1ˆ ˆ( ) ( sin )

4

qE da r r d d r q

r

The flux through any surface enclosing the charge is According to the principle of superposition, the total field is the sum of all the individual fields:

0q

E da

01 1

1( ) ( )

n n

i ii i

E da E da q

,

2.2.1 (3)

E

A charge outside the surface would contribute nothing to thetotal flux,since its field lines go in one side and out other.

1

n

ii

E E

Turn integral form into a differential one , by applying the divergence theorem

0 0

( )

1 ( )

1

surface volume

volumeenc

E da E d

Q d

0

1E

Gauss’s law in differential form

2.2.1 (4)

Gauss’s Law in integral form 0

1encE da Q

2.2.2 The Divergence of E

20

ˆ1( ) ( )

4all space

RE r r d

R

,R r r

The r-dependence is contained in

20

ˆ1( ) ( )

4

RE r d

R

From3

2

ˆ( ) 4 ( )

R

RR

Thus

3

0 0

1 14 ( ) ( ) ( )

4E r r r d r

Calculate the divergence of E directly

R

2.2.3 Application of Gauss’s Law

0

1enc

surface

E da Q

surface

E da E da

(b)E is constant over the Gaussian surface

24surface surface

E da E da E r

Thus 2

0

14E r q

20

1ˆ

4

qE r

r

(a) E point radially outward ,as does

Sol:

Example 2.2 Find the field outside a uniformly charged sphere of radius

ad

a

1. Spherical symmetry. Make your Gaussian surface a concentric sphere (Fig 2.18)2. Cylindrical symmetry. Make your Gaussian surface a coaxial cylinder (Fig 2.19)3. Plane symmetry. Use a Gaussian surface a coaxial the surface (Fig 2.20)

2.2.3 (2)

Example 2.3 Find the electric field inside the cylinder which contains charge density as Solution:

kr

0

1enc

surface

E da Q

The enclosed charge is

2 30

2( )( ) 2

3

rencQ d kr r dr d dz k l r dr klr

2E da E da E da E rl

3

0

1 22

3E rl klr

2

0

1ˆ

3E kr r

thus

2.2.3 (3)

(by symmetry)

Example 2.4 An infinite plane carries a uniform surface charge. Find its electric field.

Solution: Draw a ”Gaussian pillbox Apply Gauss’s law to this surface

0

1

surfaceencE da Q

By symmetry, E points away from the planethus, the top and bottom surfaces yields

2E da A E

(the sides contribute nothing)

0

12A E A

0ˆ

2E n

2.2.3 (4)

Example 2.5 Two infinite parallel planes carry equal but opposite uniform charge densities .Find the field in each of the three regions.

2.2.3 (5)

Solution: The field is (σ/ε0 ), and points to the right, between the plane elsewhere it is zero.

0( )2

0( )2

0( )2

0

( )2

0( )2

0

( )2

2.2.4 The Curl of EA point charge at the origin. If we calculate the line integral of this field from some point to some other point b :

b

aE dl

20 0 0

1 1 1

4 4 4

b

a

rb b

a a a br

q q q qE dl dr

r r rr

This integral is independent of path. It depends on the two end points

0E dl

( if ra = rb )Apply by Stokes’ theorem, 0E

In spherical coordinate

ˆ ˆˆ sindl dr r rd r d

20

1

4

qE dl dr

r

a

2.2.4

The principle of superposition

1 2E E E

1 2 1 2( ) ( ) ( ) 0E E E E E

so

must hold for any static charge distribution whatever.

0E

2.3 Electric potential

2.3.1 Introduction to Potential

2.3.2 Comments on Potential

2.3.3 Poisson’s Equation and Laplace’s Equation

2.3.4 The Potential of a Localized Charge Distribution

2.3.5 Electrostatic Boundary Conditions

2.3.1 introduction to potential

Any vector whose curl is zero is equal to the gradient of somescalar. We define a function:

Where is some standard reference point ; V depends only on the point P. V is called the electric potential.

The fundamental theorem for gradients

( ) ( ) ( )b

aV b V a V dl

( )b b

a aV dl E dl

so

( )

pV p E dl

( ) ( )b a b

aV b V a E dl E dl E dl

E V

2.3.2 Comments on potential

(1)The name Potential is not potential energy

F qE q V

U F X

V : Joule/coulomb U : Joule

(2)Advantage of the potential formulation V is a scalar function, but E is a vector quantity

( )V r ˆ ˆ ˆx y zE E x E y E z

If you know V, you can easily get E: .E V

0E

so

2.3.2 (2)

, ,x y zE E E are not independent functions

, ,y yx xz zE EE EE E

y x y z z x

2

y z x z y x z x y x z y x y z y x z z x x zE E E E E E E E

1 3

2 1 3

Adding a constant to V will not affect the potential difference between two point:

(3)The reference point Changing the reference point amounts to adds a constant tothe potential

( ) ( )p p

V p E dl E dl E dl K V p

(Where K is a constant)

2.3.2 (3)

( ) ( ) ( ) ( )V b V a V b V a

Since the derivative of a constant is zero:For the different V, the field E remains the same.

Ordinarily we set

V V

( ) 0V

2.3.2 (4)

(4)Potential obeys the superposition principle

1 2F F F F QE

1 2E E E

Dividing through by Q

,

Integrating from the common reference point to p ,

1 2V V V (5)Unit of potential Volt=Joule/Coulomb

:

: :qV

XF X

Vq

F newton F x Joule

F qE q V

Joule/Coulomb

solution:

0 0

1 14 4

rr q q

V r E dr r

for r>R:

for r<R:

0inE

201

4ˆq

out rE r

0

14

qV r V R

R 0

14

qV R

R

0inE

2.3.2 (5)

Example 2.6 Find the potential inside and outside a spherical shell of radius R, which carries a uniform surface charge (the total charge is q).

_

2.3.3 Poisson’s Eq. & Laplace’s Eq.

0

2E V V

E V

Laplace’s eq.0

Poisson’s Eq.0

2V

2 0V

pR r r

i i pR r r 0 1

1( )4

ni

ii

qV P

R

2.3.4 The Potential of a Localized Charge Distribution

• Potential for a point charge

• Potential for a collection of charge

E V r

V V Edr 0V

20 0 0

1 1 1( )4 4 4

rr q q q

V r drr rr

R

Ri

0

1( )4

qV P

R

• Potential of a continuous distribution for volume charge for a line charge for a surface charge

0

1( )4

V P dR

0

1( )4

V P dR

0

1( )4

V P daR

q d q d q da

20

1 ˆ( )4

rE P dR

20

ˆ14

( ) r dR

E P

20

ˆ14

( ) r daR

E P

• Corresponding electric field 2

ˆ1 RR R

[ ]

2.3.4 (2)

2.3.4 (3)Example 2.7 Find the potential of a uniformly charged spherical shell of radius R.

0

1( ) ,

4V r da

r

r2 2 2 2 cosr R z Rz

2

2 2

2

2 20

2 2 2

0

2 2 2 2

2 2

sin4 ( )

2 cos

sin2

2 cos

12 2 cos

22 2

2( ) ( )

( )

( )

[ ]

R d dV z

R z Rz

R dR z Rz

R R z RzRz

RR z Rz R z Rz

zR

R z R zz

Solution:

2

0 0

0 0

( ) [( ) ( )] ,2

( ) [( ) ( )] ,2

R RV z R z z R outsidez z

R RV z R z R z insidez

2.3.4 (4)

2.3.5 Electrostatic Boundary Condition

Electrostatic problem

The above equations are differential or integral. For a unique solution, we need boundary conditions. (e.q. , V()=0 )(boundary value problem. Dynamics: initial value problem.)

•Superposition

•Coulomb law

2.3.5 (2)

B.C. at surface with charge

0 0

encQ A

surface

E da

above belowE A E A

0above belowE E

=

∥:

0E dl

0E

E‖ above= E‖ below

+∥0

ˆabove belowE E n

:

Potential B.C.

2.3.5 (3)

00

babove below a

abV V E d

0ˆabove belowE E n

E V

ˆV V nn

0above belown nV V

above belowV V

2.4 Work and Energy in Electrostatics 2.4.1 The Work Done in Moving a Charge

2.4.2 The Energy of a Point Charge Distribution

2.4.3 The Energy of a Continuous Charge Distribution

2.4.4 Comments on Electrostatic Energy

2.4.1 The Work Done in Moving a charge

A test charge Q feels a force Q E

(from the stationary source charges).

To move this test charge, we have to apply a force

conservative

F QE

The total work we do is

b b

a aW F d Q E d Q V b V a

WV b V aQ

So, bring a charge from to P, the work we do is

( ) ( ) ( )W Q V P V QV P

( ) 0V

2.4.2 The Energy of a Point Charge Distribution

It takes no work to bring in first charges

1 0W

Work needed to bring in q2 is :

1 12 2 2

0 12 0 12

1 1[ ] ( )4 4

q qW q q

R R

1 2 1 23 3 3

0 13 0 23 0 13 23

1 1 1[ ] ( )4 4 4

q q q qW q q

R R R R

Work needed to bring in q3 is :

Work needed to bring in q4 is :

31 24 4

0 14 24 34

1 [ ]4

qq qW q

R R R

for q1

2.4.2

Total work

W=W1+ W2+ W3 +W4

1 3 2 3 3 41 2 1 4 2 4

0 12 13 23 14 24 34

14

q q q q q qq q q q q qR R R R R R

01

14

nj

iijj

qV P

R

j i

0 1 1

14

n ni j

iji j

q q

R

ij jiR R

01 1

1 12 4

n nj

iiji j

qq

R

0 1 1

18

n ni j

iji j

q q

R

1

12

n

i ii

W q V P

Dose not include the first charge

j i j i

2.4.3 The Energy of a Continuous Charge Distribution

0ρ E

q d

Volume charge density

0 0τ2 2

E Vd EV d E V d

surface

VE da

20 ( )2

surface volume

VE da E d

F.T.forE

1τ

2W Vd

surface 20

τ2

all space

W E d

EV E V E V

2.4.3

0

1

4

qV

R24q R Sol.1 ：

2

0 0

1 1 1 1 1

2 2 2 4 8

q q qW Vd Vda da

A R R

Example 2.8 Find the energy of a uniformly charged spherical shell of total charge q and radius R

Sol.2 ： 20

1ˆ

4

qE r

r

22

2 40(4 )

qE

r

2

2 20 02 4

1sin

2 2 (4 )

outer

spaceo

qW E d r d d dr

r

2 2

2 200

1 12 2

832 R

q qdr

Rr

2.4.4 Comments on Electrostatic Energy

1 A perplexing inconsistent

20 Energy 02 all space

W E d

1

1( )

2

n

i ii

W q V P

0 1 1

1

4

n ni j

iji jj i

q qW

r

or

Only for point charge 2

1 20 12

1, 0

4

qq q q q W

r

1 2( )q q

Energy 0 or 0

and are attractive

20 τ2 all space

W E d

is more complete

the energy of a point charge itself,

22

202 4 2000

1( ) sin θ θ φ

8 π2(4 )

qqW r d d dr dr

r r

1

1

2

n

i ii

W q V P

, ( )iV P does not include iq

1

2W Vd is the full potential

2.4.4 (2)

There is no distinction for a continuous distribution,because 0

( ) 0d

d p

( )V P,

(2)Where is the energy stored? In charge or in field ?

Both are fine in ES. But,it is useful to regard the energy as being stored in the field at a density

2

0 2

E Energy per unit volume

2.4.4 (3)

(3)The superposition principle,not for ES energy

201 12

W E d 20

2 2 τ2

W E d

20

1 2( )2totW E E d

1 2 0 1 2( )W W E E d

2 201 2 1 2( 2 )

2E E E E d

2.5 Conductors

2.5.1 Basic Properties of Conductors

2.5.2 Induced Charges

2.5.3 The Surface Charge on a Conductor; the Force on a Surface Charge

2.5.4 Capacitors

2.5.1 Basic Properties of Conductors

e are free to move in a conductor

E

otherwise, the free charges that produce

will move to make 0E

inside a conductor

( ) ( ) 0

( ) ( )

b

aV b V a E dl

V b V a

0E

0 0E

0E

inside a conductor(1)

(2) ρ = 0 inside a conductor

(3) Any net charge resides on the surface

(4) V is constant, throughout a conductor.

(5) is perpendicular to the surface, just outside a conductor. Otherwise, will move the free charge to make in terms of energy, free charges staying on the surface have a minimum energy.

uniform

0inE 0inW

2

0

1

8

qEnergy

R

uniform

0inE 0Win

2

0

3

20

qEnergy

R

2.5.1 (2)

E

E

0E

0

0in

enc induced

induecd

E

Q q q

q q

Example : A point charge q at the center of a spherical conducting shell. How much induced charge will accumulate there?

charge conservation2 24 4b ab a

21

4bq

b

2

0

4

in

a

E

a q

21

4aq

a

q induced

Solution :

2.5.1 (3)

2.5.2 Induced Charge

Example 2.9 Within the cavity is a charge +q. What is the field outside the sphere?

-q distributes to shield q and to make 0inE i.e., surfaceV constant

from charge conservation and symmetry.+q uniformly distributes at the surface

20

1ˆ( )

4

qE P r

r

a,b are arbitrary chosen

0

0b

a

E dl

E dl

0incavityE

a Faraday cage can shield out stray E

2.5.3 The Surface Charge on a Conductor

0ˆabove belowE E n

0 or 0in belowE E

0ˆaboveV E n

0

V

n

if we know V, we can get

or

why the average?

0

ˆ2above otherE E k

0

ˆ2below otherE E k

1( )

2other above below averageE E E E 0

electrostatic pressure2

20

02 2P E

In case of a conductor2

0

1ˆ

2 2f E nabove

2.5.3

1( )

2average above belowf E E E

Force on a surface charge,

2.5.4 Capacitors

Consider 2 conductors (Fig 2.53)

The potential difference

V V V E dl

20

ˆ1

4

rE d

r

Define the ratio between Q and V to be capacitanceQ

CV

a geometrical quantity

in mks 1 farad(F)= 1 Coulomb / volt

inconveniently large ; 6

12

10 :

10 :

F microfarad

F picofarad

Q E V ��������������

double double double double

(V is constant.)

Example 2.10 Find the capacitance of a “parallel-plate capacitor”?

0 0

QE

A

0

QV E d d

A

0AC

d

2.5.4 (2)

+Q

-QAd

Solution:

Example 2.11 Find capacitance of two concentric spherical shellswith radii a and b .

20

1ˆ

4

QE r

r

20 0

1 1 1

4 4

a a

b b

Q QV E dl dr

a br

04Q ab

CV b a

-Q+Q

2.5.4 (3)

Solution:

The work to charge up a capacitor

( )q

dW Vdq dqC

22

0

1 1 1( )

2 2 2

Q QqW dq QV CV

C C

2.5.4 (4)