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?
r r
r r
The exerted force F (r ) is used to opposite the electric force QE (r ) .
The work can be obtained via integration if the force is conservative
r
r
r
r
b r
b
b
r
r r
r r
r r
r
dV (r )
W = ∫ F ⋅ dl = −Q ∫ E (r ) ⋅ dl = Q ∫ ∇V (r ) ⋅ dl = Q ∫
dl = Q V b − V (a )
r
r
r
r
dl
a
a
a
a
r
b
[()
]
r
r
If we move charge from far away to r , the work done is W = Q[V (r ) − V (∞ )] .
r
If we set reference point at infinity as zero, W = QV (r ) .
2.4.2 The Energy of a Point Charge Distribution
The first charge q1 is moving with no work since there is no electric field. (Is that
right?)
r
Work done for moving the second charge q2 at a place r2 is:
1 q1q2
1 q1q2
W=
Wtotal =
4πε 0 Π12
4πε 0 Π12
Work done for moving the third charge in the space is:
1 q1q3
1 q2 q3
1 q1q2
1 q1q3
1 q2 q3
W=
+
Wtotal =
+
+
4πε 0 Π13 4πε 0 Π 23
4πε 0 Π12 4πε 0 Π13 4πε 0 Π 23
For the fourth charge:
1 q1q4
1 q2 q4
1 q3q4
W=
+
+
4πε 0 Π14 4πε 0 Π 24 4πε 0 Π 34
1 q1q2
1 q1q3
1 q1q4
1 q2 q3
1 q2 q4
1 q3q4
+
+
+
+
+
4πε 0 Π12 4πε 0 Π13 4πε 0 Π14 4πε 0 Π 23 4πε 0 Π 24 4πε 0 Π 34
1 n n qi q j 1 1 n n qi q j
=
For n charges the total work is W =
∑∑
∑∑
4πε 0 i =1 j ≠ i , j < i Π ij 2 4πε 0 i =1 j =1, j ≠ i Π ij
Wtotal =
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.
n
qj 1 n
r
1 n
1
W = ∑ qi
= ∑ qiV (ri )
∑
2 i =1 4πε 0 j =1, j ≠ i Π ij 2 i =1
n
r
Move in charge one by one: W = ∑ qiV (ri )due _ to _ ch arg e _ j < i
i =1
Assume all charge exists except itself: W =
r
1 n
qiV (ri )ch arg es _ except _ i
∑
2 i =1
Example: Find the energy required to assemble a uniform sphere of charge of radius
b and a volume charge of ρ .
Move in shells of charge to assemble the sphere.
4π 3
r ρ
ρr 2
W = ∫ VdQ , V (r ) = 3
, dQ = 4πr 2 drρ
=
4πε 0 r
3ε 0
2


b
2
2
5 
4πρ 5 4πb
ρr
Q 
3Q 2
2


4πr ρdr =
W =∫
b =
=
3ε 0
15ε 0
15ε 0  4π b3 
20πε 0b
0



 3
2.4.3 The Energy of a Continuous Charge Distribution
Assumes all charges already exist the potential experiences by the volume charge ρ
r
r
r
r
1
1
is V (r )all _ other _ ch arg es and the energy is: W = ∑ qiV (ri ) = ∫ ρ (r ')V (r ')dτ ' .
2
2
r
r r
r r
r
ρ (r ')
∇'⋅E (r ') =
Æ ρ (r ') = ε 0∇'⋅E (r ')
ε0
[ (
ε0
]
)
r r
r r
r
r
r
r
ε0
(
)
(
)
(
)
(
)
∇
'
⋅
E
r
'
V
r
'
d
τ
'
=
∇
'
⋅
E
r
'
V
r
'
−
E
⋅ ∇V (r ') dτ '
∫
∫
2
2
r 2
r r r
ε
ε
W = 0 ∫ E (r ') dτ ' + 0 ∫ V (r ')E (r ')da'
2
2 surface
W=
The surface integral goes to zero at infinity since V ∝
1
1
, E ∝ 2 , and
r
r
surface _ area ∝ r 2 .
W=
ε0
r
E (r ') dτ '
2∫
2
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 q
and radius R.
Method 1: W =
1
1
q
q
q2
VdQ
,
=
=
W
da
2∫
2 ∫ 4πε 0 R 4πR 2
8πε 0 R
∞
ε0 
2

q2
2

4
π
Method 2: W = ∫ E dτ , W = ∫ 
=
r
dr
2
2 R  4πε 0 r 2 
8πε 0 R
ε0
2
q
Example: Find the energy required to assemble a uniform sphere of charge of radius
b and a volume charge of ρ .
Use the method: W =
ε0
2
∫ E dτ , find the electric field firstly
2
4π 3
bρ
b3 ρ
r > b, E = 3
=
4πε 0 r 2 3ε 0 r 2
4π 3
r ρ
ρ
r <b, E = 3 2 =
r
4πε 0 r
3ε 0
W=
ε 0 ∞ b6 ρ 2
ε0 b ρ 2 2
2
4
r
dr
+
r 4πr 2 dr
π
2 4
2
∫
∫
2 b 9ε 0 r
2 0 9ε 0
W=
ε 0  4πb 6 ρ 2  1  4πρ 2 b5  4π 2 5
ρb
 +

=
2  9ε 02  b  9ε 02 5  15ε 0
Calculate the energy:
1.
r
move in charge one by one: W = ∑ qiV (ri )
2.
all charges already settle down: W =
r
1
1
qiV (ri ) = ∫ VdQ (integrate the
∑
2
2
volume having charges)
3.
by electric field: W =
ε0
E dτ
2∫
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: W = ∫
∞
ε0 
2

C
 4πr 2 dr = = ∞
E dτ = ∫ 
2 
2
2 0  4πε 0 r 
0
ε0
2
q
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
ε0
2
r2
E (r ) .
In electrostatics, we can say that it is stored in the charge with a volume density
1 r r
ρ (r )V (r )
2
(iii) The superposition principle?
The electrostatic energy is quadratic in the fields?
The electric potential is linear in the fields?
r r 2
r r
ε
ε
ε
W = 0 ∫ E 2 dτ = 0 ∫ E1 + E2 dτ = 0 ∫ E12 + E22 + 2 E1 ⋅ E2 dτ
2
2
2
r r
= W1 + W2 + ε 0 ∫ E1 ⋅ E2 dτ
(
)
(
)
Is the concept of electric potential similar to that of potential energy?
E ∝ Q , F ∝ Q2 , V ∝ Q , W ∝ Q2
Is the concept of electric field the same as that of force?
Q ∝ E , W ∝ Q2 ∝ E 2 ? F ∝ Q2 ∝ E 2
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
r
(i) E = 0 inside a conductor.
If there were any field inside a conductor, those free charges would move.
Why is the charge always on the surface of a
+e
conductor? (the electric field move the
-e -e
+e +e
free charge to the boundary and the new
-e
+e
distributed charge generate a field to cancel
-e
the external field)
(ii) ρ = 0 inside a conductor.
No boundary inside a conductor. (It’s possible to trap electron inside a conductor?)
r ρ
∇⋅E =
Å Æ the linear relation between the charge and its field?
ε0
r
If E = 0 , so also is ρ .
(iii) Any net charge resides on the surface.
Boundary?
(iv) A conductor is an equipotential.
r
It’s equipotential since E = 0 inside a conductor.
r r
That is any two point a , b inside the conductor, there potential difference
()
r
r
r
a
r
a
()
b r
b
r
r
r
r
r
r
V b − V (a ) = − ∫ E ⋅ dl = − ∫ 0 ⋅ dl = 0 Æ V b = V (a )
r
(v) E is perpendicular to the surface, just outside a conductor.
The tangential field will move the free electron to cancel
the field itself. The electric field is perpendicular to the
surface. Boundary problem?
tangential field
The charge on a conductor will seek the configuration that minimizes its potential
energy.
The energy of a sphere with surface charge is
volume charge is
1 q2
while that of a sphere with
8πε 0 R
3 q2
.
20πε 0 R
2.5.2 Induced Charges
q
-
+
-
+
+
+
+
r
E ≠ 0 -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
V=
1
dq '
4πε 0
Q
∫ rr − rr ' = 4πε R
Æ What’s the capacity of something to storing charges
0
per volt? Capacitance: C =
Q
.
V
+Q
d
-Q
Method 1 to explain the electrostatic pressure:
Move additionally infinitesimal dQ to the condenser, you need energy
dU = dW = VdQ =
 Q2 
Q2
Q
 Æ U =
dQ = d 
2C
C
 2C 
σ
σd Qd
Q Aε 0
dz =
=
, C= =
ε
ε 0 Aε 0
V
d
0 0
d
V = ∆V = ∫
The change of energy with variation of d:
∆U =
∆U
Q 2  1  Q 2 ∆d
Q2
F σ2
Æ F=
=
Æ P= =
∆  =
∆d 2ε 0 A
2  C  2 Aε 0
A 2ε 0
Method 2 to explain the electrostatic pressure:
F=
σ σ /2 σ σ /2 σ2
∗
+ ∗
=
2 ε0
2 ε0
2ε 0
σ
2
material
Method 3: average electric field
The charge per unit area is σ and the average electric field is
r
r
r
r
1 r
Eavg = Eabove + Ebelow , so the force per unit area is f = σ ∗ Eavg
2
(
)
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.
W =U = ?, F = −
W=
dU
=?
dx
x
Q
∂W
Q2
ε S
Q2
, C=
= 0 , F =−
=−
Q
2C
x
∂x
2ε 0 S
x
ε 0S
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+ r
r
Vdiff = V+ − V− = − ∫ E ⋅ dl
r−
2.
r
E and V are proportional to Q.
3.
The constant of proportionality is called the capacitance of the arrangement.
r r
E (r ) =
ˆ
Π
dq'
4πε 0 ∫ Π 2
1
Q
(capacity of storing charges per volt)
V
C is measured in farads (F). Usually we use µF
and pF .
C=
4.
Capacitance is purely geometrical quantity,
determined by the sizes, shapes, and separation of the
two conductors. Æ SET
The capacitance of a single conductor Æ 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.
r+ Q
r
 Q 
σ
Qd
Q
zˆ , V+ − V− = − ∫  −
zˆ dz =
E = − zˆ = −
ε0
ε0 A
Aε 0
Aε 0 
r−Q 
C=
Q Aε 0
=
V
d
Example: capacitor of two concentric metal shells
Find the capacitance of two concentric spherical metal shells,
with radii a and b.
a
r
Q rˆ
Q  1 1
Q rˆ
, V = −∫ −
⋅ rˆdr =
E=−
− + 
2
2
4πε 0  a b 
4πε 0 r
4πε 0 r
b
C=
+Q, A
Q
ab
= 4πε 0
V
a−b
Work for charging up a capacitor:
dU = dW = VdQ =
 Q2 
Q 2 CV 2
Q
 Æ W =
=
dQ = d 
2C
2
C
 2C 
Again, remember the relationship: Q ∝ V , W ∝ Q 2 ∝ V 2
b, -Q
a
z
Remind: A Cylindrical Problem
∞
V=
∫ 4πε
−∞
λdx
x +z
2
0
2
=?
Apply Gauss’s law: E =
r
z
1 λL
rˆ
2πrL ε 0
λ
λ
s
dr = −
ln 
2πrε 0
2πε 0  a 
a
s
−λ
V = −∫
− λ 
ln
V =−
2πε 0 


λ
V=
ln
2πε 0 

(x − (− a ))
2
a'

+ y 
λ
ln
−
 2πε 0 


2
(x − a ) + y 

a'

(x + a )2 + y 2  λ  (x + a )2 + y 2 
ln
=
2
(x − a )2 + y 2  πε 0  (x − a ) + y 2 
Exercise: 35, 38, 39, 46, 47, 48
2
2
z
a
+λ