2.4 Work and Energy in Electrostatics

2.4.1 The Work Done to Move a Charge How much work is done for moving a charge in an electric field?

The exerted force ( )rF rr is used to opposite the electric force ( )rEQ rr .

The work can be obtained via integration if the force is conservative

( ) ( ) ( ) ( ) ( )[ ]aVbVQdldl

rdVQldrVQldrEQldFWb

a

b

a

b

a

b

a

rrrrrrrrrrr

r

r

r

r

r

r

r−==⋅∇=⋅−=⋅= ∫∫∫∫

If we move charge from far away to rr , the work done is ( ) ( )[ ]∞−= VrVQW r . If we set reference point at infinity as zero, ( )rQVW r

= .

2.4.2 The Energy of a Point Charge Distribution The first charge is moving with no work since there is no electric field. (Is that right?)

1q

Work done for moving the second charge at a place 2q 2rr is:

12

21

041

Π=

qqWπε

12

21

041

Π=

qqtotal πε

W

Work done for moving the third charge in the space is:

23

32

013

31

0 41

41

Π+

Π=

qqqqWπεπε

23

32

013

31

012

21

0 41

41

41

Π+

Π+

Π=

qqqqqqWtotal πεπεπε

For the fourth charge:

34

43

024

42

014

41

0 41

41

41

Π+

Π+

Π=

qqqqqqWπεπεπε

34

43

024

42

023

32

014

41

013

31

012

21

0 41

41

41

41

41

41

Π+

Π+

Π+

Π+

Π+

Π=

qqqqqqqqqqqqWtotal πεπεπεπεπεπε

For n charges the total work is ∑ ∑∑ ∑= ≠== <≠ Π

=Π

=n

i

n

ijj ij

jin

i

n

ijij ij

ji qqqqW

1 ,101 ,0 41

21

41

πεπε

You don’t have to consider the order that charges are moved in the space. Just calculate the charge with its potential produced by other charge and then divide by 2.

( )∑∑ ∑== ≠=

=Π

=n

iii

n

i

n

ijj ij

ji rVq

qqW

11 ,10 21

41

21 r

πε

Move in charge one by one: ( )∑=

<=n

iijechtodueii rVqW

1_arg__

r

Assume all charge exists except itself: ( )∑=

=n

iiexcepteschii rVqW

1__arg2

1 r

Example: Find the energy required to assemble a uniform sphere of charge of radius and a volume charge of b ρ .

Move in shells of charge to assemble the sphere.

∫= VdQW , ( )0

2

0

3

343

4

ερ

πε

ρπr

r

rrV == , ρπ drrdQ 24=

bQ

b

QbbdrrrWb

0

2

2

30

55

0

2

0

2

0

2

203

3415

41544

3 πεπεπ

επρρπ

ερ

=

=== ∫

2.4.3 The Energy of a Continuous Charge Distribution Assumes all charges already exist the potential experiences by the volume charge ρ

is ( ) eschotherallrV arg__r and the energy is: ( ) ( ) ( )∫∑ == '''

21

21 τρ drVrrVqW ii

rrr .

( ) ( )0

'''ερ rrEr

rr=⋅∇ ( ) ( )''' 0 rEr rrr

⋅∇= ερ

( ) ( ) ( ) ( )( ) ( )[ ]∫∫ ∇⋅−⋅∇=⋅∇= '''''2

''''2

00 τετε drVErVrEdrVrEW rrrrrrrr

( ) ( ) ( )∫∫ +=surface

darErVdrEW '''2

''2

020 rrrr ετε

The surface integral goes to zero at infinity since r1

∝V , 21r

E ∝ , and

. 2_ rareasurface ∝

( )∫= '2

20 τ 'ε drEW r Remember to integrate over all space.

Calculate the electric field all over the space at first. Example: Find the energy of a uniformly charged spherical shell of total charge and radius R.

q

Method 1: ∫= VdQW21 ,

Rqda

Rq

RqW

0

2

20 8442

1πεππε

== ∫

Method 2: ∫= τε dEW 20

2,

Rqdrr

rqW

R 0

22

2

20

0

84

42 πεπ

πεε

=

= ∫

∞

Example: Find the energy required to assemble a uniform sphere of charge of radius and a volume charge of b ρ .

Use the method: ∫= τε dEW 20

2, find the electric field firstly

br > , 20

3

20

3

343

4

rb

r

bE

ερ

πε

ρπ

== br < , rr

rE

02

0

3

343

4

ερ

πε

ρπ

==

∫∫ +=∞ b

b

drrrdrrr

bW0

2220

202

420

260 4

924

92π

ερεπ

ερε

52

0

5

20

2

20

260

154

5941

94

2bb

bbW ρ

επ

επρ

ερπε

=

+

=

Calculate the energy:

1. move in charge one by one: ( )∑= ii rVqW r

2. all charges already settle down: ( ) ∫∑ == VdQrVqW ii 21

21 r (integrate the

volume having charges)

3. by electric field: ∫= τε dEW 20

2 (integrate over all space)

2.4.4 Comments on Electrostatic Energy (i) “inconsistency” (Ref: Feynman Lectures V2, 8-6.) When moving in charge one by one, we suppose the first charge do not suffer from any force and do not need work to move it. Is that right? If we calculate the electrostatic energy of a point charge by the method of electric

fields: ∞==

== ∫ ∫

∞

04

422 0

22

20

020 Cdrrr

qdEW ππε

ετε

The infinity comes from the chopping of a point charge to infinitesimal parts. We must conclude that the assumption of point charges may not be correct. Even as an electron has special distribution of charges. It is not a point charge. (ii) Where is the energy stored? In radiation theory, it is useful to regard the energy as being stored in the field, with a

volume density ( )20

2rE rε .

In electrostatics, we can say that it is stored in the charge with a volume density

( ) ( )rVr rrρ21

(iii) The superposition principle? The electrostatic energy is quadratic in the fields? The electric potential is linear in the fields?

( ) ( )

∫∫∫∫

⋅++=

⋅++=+==

τε

τετετε

dEEWW

dEEEEdEEdEW

21021

2122

21

0221

020 2222

rr

rrrr

Is the concept of electric potential similar to that of potential energy? QE ∝ , , , 2QF ∝ QV ∝ 2QW ∝

Is the concept of electric field the same as that of force? EQ ∝ , ? 22 EQW ∝∝ 22 EQF ∝∝

Exercise: 31, 32

2.5 Conductors

2.5.1 Basic Properties In an insulator, each electron is attached to a particular atom. In a conductor, electrons per atom are free to move around the material.

localized electron

free electron or band electron (i) inside a conductor. 0=E

r

If there were any field inside a conductor, those free charges would move. Why is the charge always on the surface of a conductor? (the electric field move the free charge to the boundary and the new distributed charge generate a field to cancel the external field)

-e -e -e

+e +e +e

-e +e

(ii) 0=ρ inside a conductor. No boundary inside a conductor. (It’s possible to trap electron inside a conductor?)

0ερ

=⋅∇ Er

the linear relation between the charge and its field?

If , so also is 0=Er

ρ .

(iii) Any net charge resides on the surface. Boundary? (iv) A conductor is an equipotential. It’s equipotential since inside a conductor. 0=E

r

That is any two point barr,

−= ∫b

a

r

r

inside the conductor, there potential difference

( ) ( ) 00 =⋅⋅−=− ∫b

a

ldldEaVbVr

r

rrrrr ( ) ( )aVbV rr=

(v) E

r is perpendicular to the surface, just outside a conductor.

tangential field

The tangential field will move the free electron to cancel the field itself. The electric field is perpendicular to the surface. Boundary problem? The charge on a conductor will seek the configuration that minimizes its potential energy.

The energy of a sphere with surface charge is Rq2

081πε

while that of a sphere with

volume charge is Rq2

0203πε

.

2.5.2 Induced Charges

0Er

q

-q q ≠q

+

++

++

-

- -

- -

Example: An uncharged spherical conductor centered at the origin has a cavity of some weird shape carved out of it. Somewhere within the cavity is a charge q. What’s the field outside the sphere?

q

If a cavity surrounded by conducting material is itself empty of charge, then the field within the cavity is zero.

2.5.3 Surface Charge and the Force on a Capacitor

RQ

rrdqV

00 4''

41

πεπε=

−= ∫ rr What’s the capacity of something to storing charges

per volt? Capacitance: VQC = .

-Q

+Q d

Method 1 to explain the electrostatic pressure: Move additionally infinitesimal dQ to the condenser, you need energy

====

CQddQ

CQVdQdWdU

2

2

C

QU2

2

=

000 0 εεσ

εσ

AQdddzVV

d

===∆= ∫ , d

AVQC 0ε==

The change of energy with variation of d:

0

22

21

2 εAdQ

CQU ∆

=

∆=∆

AQ

dUF

0

2

2ε=

∆∆

= 0

2

2εσ

==AFP

2σMethod 2 to explain the electrostatic pressure:

0

2

00 22/

22/

2 εσ

εσσ

εσσ

=∗+∗=F material

Method 3: average electric field The charge per unit area is σ and the average electric field is

( belowaboveavg EEE )rrr+=

21 , so the force per unit area is avgEf

rr∗=σ

Example: A metal sphere of radius R carries a total charge Q. What is the force of repulsion between the northern ‘hemisphere’ and the southern ‘hemisphere’?

Example: Determine the force on the conducting plates of a charged parallel plate capacitor. The plates have an area S and are separated in air by a distance x.

x ?==UW , ?=−=dxdUF

CQW2

2

= , xS

xS

QQC 0

0

ε

ε

== , S

QxWF

0

2

2ε−=

∂∂

−=

2.5.4 Capacitors What’s a capacitor? Suppose we have two conductors, and we put charge +Q on one and –Q on the other. Properties:

1. V is constant over a conductor. ∫+

−−+ ⋅−=−=

r

diff ldEVVVrr

r

2. Er

and V are proportional to Q. ( ) ∫ ΠΠ

= 'ˆ

41

20

dqrEπε

rr

3. The constant of proportionality is called the capacitance of the arrangement.

VQC = (capacity of storing charges per volt)

4. C is measured in farads (F). Usually we use Fµ and . pF

Capacitance is purely geometrical quantity, determined by the sizes, shapes, and separation of the two conductors. SET The capacitance of a single conductor the second conductor is a shell with infinite radius. Aluminum Electrolytic Capacitors Ceramic Capacitors Film Capacitors Tantalum Capacitors SET: (Ref: http://www.glue.umd.edu/~bekane/QC/QC@UMD's_LPS_Single%20Electron%20Transistor.htm) Two electrons are detected in a singlet state by conductance

measurements on a single electron transistor (SET). Applying

a change in the gate voltage between the p-doped Si substrate

and the center Al island ionizes one of the two electrons of a

Te double donor in Si. As the electron probability density

crowds against the barrier, the discrete energy levels of the

island shift. When the Fermi Energy of the island matches the

Fermi Energy of the source and drain, current can flow by

tunneling across the potential barriers between the source,

island, and drain. The same change in gate voltage applied to a

triplet does not realize the same change in conductivity

because the triplet is already ionized at the lower gate voltage.

Consult experts: C. D. Chen at Institute of Physics, Academic Sinica, and Watson Kuo at Dept. Physics, NCHU. Example: parallel plate capacitor Find the capacitance of a “parallel-plate capacitor” consisting of two metal surfaces of area A held a distance d apart.

+Q, A z

AQzE ˆˆ

00 εεσ

−=−=r

, A

QddzzAQVV

Q

Q

r

r 00

ˆεε

=

−−=− ∫

+

−

−+

dA

VQC 0ε==

b, -Q a

Example: capacitor of two concentric metal shells Find the capacitance of two concentric spherical metal shells, with radii a and b.

20

ˆ4 r

rQEπε

−=r

,

+−=⋅−−= ∫ ba

QdrrrrQV

a

b

114

ˆˆ4 0

20 πεπε

baab

VQC

−== 04πε

Work for charging up a capacitor:

====

CQddQ

CQVdQdWdU

2

2

22

22 CVC

QW ==

Again, remember the relationship: VQ ∝ , 22 VQW ∝∝

z

r

z Remind: A Cylindrical Problem

?4 22

0

=+

= ∫∞

∞− zxdxV

πελ

Apply Gauss’s law: rLrL

E ˆ2

10ε

λπ

=

−=−= ∫ a

sdrr

Vs

a

ln22 00 πελ

επλ

( )( ) ( )

+−−

+−−−−=

'ln

2'ln

2

22

0

22

0 ayax

ayaxV

πελ

πελ

λ− λ+

z

a

( )( )

( )( )

+−++

=

+−

++= 22

22

022

22

0

lnln2 yax

yaxyax

yaxVπελ

πελ

Exercise: 35, 38, 39, 46, 47, 48