Electromagnetism
Physics 15b
Lecture #7
Capacitance
Purcell 3.5–3.8
What We Did Last Time
Studied electric field near and around conductors
Just outside a conductor surface,
  E = 0 in an empty space inside a
conductor

Concentric spheres
+ +
+
+
+
+ +
+
E
E=0
+
+
+
area A
+
+++
+
+
+++
+
E = 4πσ n̂
+
Er
φ
+
+
r
r
1
Today’s Goals
Study a system with a conductive plane

Demonstrate an image charge technique
Define capacitance
First, for an isolated conductor w.r.t. infinity
  Second, for a pair of conductors
  You’ll work on this in Lab 1
  Dimension and unit
  Energy stored in a capacitor

Discuss multi-conductor systems
Generalize capacitance into a matrix
  Invert the matrix to prove another form
of the uniqueness theorem

Conductive Plane
A point charge Q is placed above a large conductive plane
It attracts negative charge distribution
on the plane’s surface
  Field lines from Q terminate on the
induced negative charge distribution

+Q
h
Need to solve Laplace’s Equation
above the plane, with a singularity
at Q and a flat equipotential surface

Generally a hard problem, but there is
a trick for this one
−
−
− −
−
2
Image Charge Technique
Consider two charges +Q and −Q separated by 2h
Coulomb field x 2 — Easy enough to calculate
  E is vertical on the mid-plane
  The mid-plane is therefore an equipotential

z
+Q
h
The field above the mid-plane satisfies
the boundary condition of the conductive
plane problem

h
By the Uniqueness Theorem, it’s the solution
This technique works in a problem with
one (infinite) conductive plane

−Q
The −Q charge is called the “image charge” of the original
Surface Charge
z
E field on the mid-plane points down
−Q
−2Qh
E z (z = 0) = 2 × 2
cos θ = 2
2
r +h
(r + h 2 )3 2

+Q
It’s related to the surface charge density
r
E
−Qh
ρ(r ) = z =
4π 2π (r 2 + h 2 )3 2

∫
∞
0
h
−Q
If we integrate over the entire surface, we get
∞
⎡ Qh ⎤
−Qhr
ρ(r )2π rdr = ∫
dr
=
⎢ 2
⎥
0 (r 2 + h 2 )3 2
2
⎣ r + h ⎦0
= −Q
∞
h
θ
r
−
−
− −
−
3
Does This Make Sense?
Total charge “induced” on the plane is –Q
Where did the corresponding positive charge go?
Didn’t we start with a neutral conductor?
−
  Culprit: the plane is “infinite”

−
− −
−
If the plane were finite, we’d see +Q
distributed around the edge and on the back

Exact result known for a disk (see Purcell) but beyond 15b
+ +
+ +
− − −−−−− − −
+
+ +
+ +
Capacitance
A conductor with charge Q has a potential φ0 w.r.t. infinity
surface
φ0 = −
∫
E ⋅ds ∝ Q
because E is proportional to Q
∞
We define capacitance by the ratio C ≡
Q
φ0
It is determined by the shape of the conductor

Example: a spherical conductor with radius a
E=
Q
r̂ (for r > a)
r2
φ0 =
Q
a
C=a
Dimension of capacitance = (length)
Unit of capacitance = cm
4
Capacitance
Capacitance can also be defined for a pair of conductors

This is in fact much more common
Give two conductors +Q and −Q and
measure the potential difference
+Q
−Q
1
φ1 − φ2 = − ∫ E ⋅ d s ∝ Q
2
Capacitance is defined by C =
Q
φ1 − φ2
Most useful case: two plates with a small gap in between
area A
s
Parallel-Plate Capacitor
If A is large (s is small), then this is similar to the double
infinite sheets of charge we saw in Lecture 3

E is uniform between the plates
E = 4πσ =

Potential difference is
φ1 − φ2 = Es =

4π Q
A
Capacitance is
4π Qs
A
C=
+Q/A
E
−Q/A
Q
A
=
φ1 − φ2 4π s
5
Real World Capacitors
Capacitors appear everywhere in electronics


Computer memory (dynamic RAM) is an array of tiny capacitors
Will do more on this later
In electronics, we must use SI

Charge is in coulomb (C), potential is in volt (V)
SI unit for capacitance is farad (F) =
coulomb (C)
volt (V)
Conversion from CGS: 1 farad = 9 x 1011 cm
  Since 1 farad is such a large unit, we often use

1 µF = 10 −6 F

1 pF = 10 −12 F
Ex: a spherical conductor with a 10 cm radius has a capacitance of
10 cm = 11 pF
Energy in a Capacitor
E field inside a parallel-plate capacitor contains energy
E2
2π Q 2s Q 2
U=∫
dV =
=
V 8π
A
2C
+Q/A
This is also how much work is
needed to “charge up” the
capacitor from 0 to Q
−Q/A
E=
4π Q
A
q
When the charge is q, the potential difference is Δφ =
C
  Work required to move a small charge dq
qdq
from the bottom plate to the top plate is dW = Δφ dq =


Integrate: W =
∫
Q
0
qdq Q 2
=
C
2C
C
Q2
Generally for any capacitor, energy U =
2C
6
Multiple Conductors
φ0 = 0
Generalize to multiple conductors

Everything is enclosed in a (possibly infinite)
boundary at φ0 = 0
φ1,Q1
If potential φ1, φ2, φ3, … of all conductors
are given, φ is uniquely determined
everywhere  Uniqueness theorem

φ2 ,Q2
φ3 ,Q3
Once φ is known, so is E, and so is σ on all
conductor surfaces
 Charges Q1, Q2, Q3, … of all conductors are determined
Superposition principle guarantees:
Qi = Ci1φ1 + Ci 2φ2 + C13φ3 +  = ∑ Cijφ j
j

Constants Cij are coefficients of capacitance
Coefficients of Capacitance
To determine Cij, set φ1 > 0 and all other φi = 0
φ0 = 0
Qi = Ci1φ1

φ1 > 0
All field lines start from 1 and end on other
conductors or the boundary B
Q1 > 0, Qi ≠1 ≤ 0, QB ≤ 0

Total charge including QB must be zero
∑ Qi + QB = 0 →
i
∑ Qi ≥ 0

φ2 = 0
i
So the coeffs. Cij must satisfy
Cii > 0, Ci ≠ j ≤ 0, and
φ3 = 0
∑C
ij
≥0
j
Also: Cij = Cji  see Purcell Problem 3.27
⎡
⎢
[Cij ] = ⎢
⎢⎣
+
−
−
+
−
−
−
−
+
⎤
⎥
⎥
⎥⎦
7
Potential Coefficients
Can we solve Qi = ∑ Cijφ j for φj?
j
Only if the [Cij] matrix is invertible, i.e, |Cij| ≠ 0
  The “shape” of [Cij] we’ve found on the last slide ensures |Cij| ≠ 0

So we can invert the matrix and get
φi = ∑ PijQ j , where P = C −1
j

Pij are called the potential coefficients
At this point we know that specifying charges Qj of all
conductors (plus the potential φ0 of the boundary) uniquely
determines the field φ everywhere

Physically “obvious” but also mathematically true
Summary
Image-charge technique for a conductive plane
Definitions of capacitance
z
+Q
Single isolated conductor
Q
Q
C≡
or
Between two conductors
φ
φ1 − φ2
  Unit: cm in CGS, farad in SI
  Parallel-plate capacitor

h

area A
s
A
C=
4π s
h
−Q
2
Q
2C
General multi-conductor system

Energy stored in a capacitor: U =
Qi = ∑ Cijφ j
j
φi = ∑ PijQ j , where P = C −1
j
8