Energy of charged cylinders
An infinitely long cylinder has radius a and is electrically charged with uniform charge density ρ.
Find
a) the electric field E(r),
b) the electric potential ϕ(r) (with the condition ϕ(0) = 0),
c) the electrostatic energy per unit length, using the volume integral
Z
1
Ues =
ρϕdτ.
(1)
2
d) Now consider two identical infinite cylinders, parallel to each other, posed at a distance d > 2a.
Find the potential energy of the system (per unit length).
e) Now try to determine the electrostatic energy of the cylinder using the integral
Z
1
′
Ues = ǫ0 E2 dτ,
(2)
2
explaining the different result with respect to that obtained in a).
1
Solution
a) For symmetry reasons the electric field is radial (in cylindrical coordinates) and dipends only on
r, the distance from the axis of the cylinder. To find the field we apply Gauss’s law to a coaxial
cylindrical surface of radius r and height h, obtaining
 ρ

r, 0 6 r 6 a,


 2ǫ0
(3)
E(r) =
2


ρa


r̂, r > a.
2ǫ0 r
b) The electric potential can be obtained by integrating the electric field obtained in a)
Z r
ϕ(r) − ϕ(r0 ) = −
E(r)dr.
(4)
r0
In the region 0 6 r 6 a, posing r0 = 0 we find
ϕin (r) = ϕin (0) −
ρr2
,
4ǫ0
(5)
where it is “natural” to assume ϕin (0) = 0, thus ϕin (a) = −ρa2 /(4ǫ0 ). In the region r > a, imposing
ϕ to be continuous at r = a we find
r
r
ρa2
1
ρa2
log
log
=−
+
.
(6)
ϕout (r) = ϕout (a) −
2ǫ0
a
2ǫ0
a
2
c) We calculate the integral (1) over a finite length h in z. In cyclindrical coordinates dτ = rdrdθdz.
Since ρ = for r > a, the integral extends on 0 < r < a and
Z
Z
1
1 2πhρ2 a 3
Ues =
ρϕdτ = −
r dr
(7)
2 cyl
2 4ǫ0 0
from which we obtain the energy per unit length
Ues = Ues /h = −
π 2 4
ρa.
16ǫ0
(8)
d) For the system of the two cylinders we might use the formula (1) to compute the electrostatic
energy but the calculation would be lengthy. Instead, we evaluate the energy as the sum of the two
identical contributions from the two cylinders given by Eq.(8) and of the interaction energy between
the two cylinders. The latter can be calculated, as a consequence of Gauss’s law, as the energy of
a charged wire of length h and linear charge density λ = πa2 ρ in the potential of a parallel, infinite
wire with the same linear charge density λ:
d
Uint (d)
πρ2 a4
1
log
Uint (d) = ϕout (d) · λh ⇒ Uint =
=−
+
.
(9)
h
2ǫ0
a
2
2
We thus obtain per unit length
Ues (pair) = 2Ues + Uint
d
1
π 2 4 πρ2 a4
log
+
= − ρa −
8ǫ0
2ǫ0
a
2
2 4
πρ a
3
d
= −
+
.
log
2ǫ0
a
4
(10)
Notice that it is not necessary to assume d ≫ 2a.
e) If we use Eq.(2), splitting the integral in the two contributions from the regions r < a and r > a
we obtain
Z
Z
1
1
2
Ues =
ǫ0
E dτ + ǫ0
E2 dτ
(11)
2
2
r>a
06r6a
2 2 Z ∞
2 Z a
1
1
ρa
ρ
1
r3 dr,
(12)
ǫ0 2πh
rdr + ǫ0 2πh
=
2
2
2ǫ0
r
2
2ǫ
0
a
0
and the first integral is divergent. This is because the two formulas (1) and (2) for the electrostatic
energy actually differ by a term
Z
Z
1
1
− ǫ0 ∇(ϕE)dτ = − ǫ0 (ϕE)n̂dS
(13)
2
2
where we used the divergence theorem and the vector identity ∇ · (AB) = B∇A + A∇ · B. The
flux integral over a coaxial cylindrical surface is divergent because dS is of order r, the electric field
decays as 1/r and the potential is proportional to log(r/a). In general the formula (2) must not be
used for a charge distribution extending to infinity as in the present case.
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