Introductory Physics Volume Two

2.5 Energy in an Electric Field 47
a wire (radius a) surrounded by a cylindrical conducting shield (radius
b) Show that the capacitance for a length L of coaxial cable is 2πɛ0L
ln(b/a) .
⊲ Problem 2.12
Consider two long, parallel, and oppositely charged wires of radius a
with their centers separated by a distance b. Assuming the charge
is distributed uniformly on the surface of each wire, show that the
capacitance per unit length of this pair of wires is
C
l = πɛ o
ln( b a − 1)
§ 2.5 Energy in an Electric Field
Let us consider how much work must be done to charge a capacitor.
Suppose that we have already moved an amount of charge q from the
negative plate to the positive plate of the capacitor, so that the electric
potential difference between the plates is ∆V = q/C. In order to move
a little bit more charge dq to the positive plate, we need to do an
amount of work
dW = dU = dq ∆V = dq q C
The total amount of work to bring the capacitor from a charge of q = 0
to a charge of q = Q is
U =
∫ Q
0
dq q C = Q2
2C
=
(C∆V )2
2C
= 1 2C(∆V )2
Theorem: Energy Stored in a Capacitor
A capacitor charged to an electric potential difference of V C stores
an amount of energy
U = 1 2 CV 2 C
This energy is stored in the electric field that has been created
between the plates. The energy density (energy per volume) is
u = energy 1
1 ɛ
volume = 2C(∆V )2
0A
2 d
=
(Ed)2 = 1
V
A d
2 ɛ 0E 2
In the intermediate step in the above equation, the properties of a parallel
plate capacitor was used, but the final result is true in general.
Theorem: Energy Density of an Electric Field
The electric field contains an amount of energy per volume u.
u = 1 2 ɛ 0E 2