Uniformly Charged Wire Outside a Grounded Cylinder—C.E. Mungan, Spring 1998
Suppose that an infinite wire carrying a uniform linear charge density λ runs parallel to an
infinitely long, grounded, conducting cylinder of radius R. The wire is located a distance a away
from the center of the cylinder, where a > R. The problem is to find the potential everywhere
outside the cylinder.
y
r
a
0
λ
x
R
We first find the potential ψτ due to the wire alone. Gauss’ law applied to a cylindrical
gaussian surface of radius r and length L concentric with a wire crossing the origin implies
q
λL
λrˆ
∫ E ⋅ dA = εenc0 ⇒ E 2π rL = ε 0 ⇒ E = 2πε 0r
λ
ln r
2πε 0
in cylindrical coordinates, where for simplicity the arbitrary constant (which sets the zero of
potential) has been chosen to be zero. Shifting the origin a distance a off the wire now gives
λ
λ
ψτ = −
ln r − axˆ = −
ln ( x − a)2 + y 2
2πε 0
2πε 0
∴ V = − ∫ E ⋅ dr = −
λ
ln(r 2 − 2 ar cosθ + a 2 ).
4πε 0
Note that, unlike for charge distributions of finite size, this potential does not tend to zero at
infinity!
Next, the potential ψS due to the cylinder alone is a solution of Laplace’s equation. In
cylindrical coordinates the potential is obviously independent of z and hence is a solution of
Laplace’s equation in plane polar coordinates. By Boas Problem 13.5.12, this can be written
most generally as
∞
D
ψ S = B0 (C0 + D0 ln r ) + ∑  Cn r n + nn  ( An sin nθ + Bn cos nθ ) .
r
n =1
Clearly at large distances from the cylinder, it looks like a line charge and hence ψ S ~ ln r as
r → ∞ ; therefore Cn = 0. Absorbing the extraneous constant B0 into C0 and D0 , and similarly Dn
into An and Bn, now leaves
∞
1
ψ S = C0 + D0 ln r + ∑ n ( An sin nθ + Bn cos nθ ).
n =1 r
=−
Assembling the solution ψ = ψ τ + ψ S , the boundary condition ψ ( R,θ , z ) = 0 leads to
∞
λ
1
a0 + ∑ ( an cos nθ + bn sin nθ ) =
ln( R2 − 2 aR cosθ + a 2 )
πε
2
4
0
n =1
=
λ
R
ln a 2 (1 − 2 k cosθ + k 2 ) where k ≡ < 1
a
4πε 0
λ
λ ∞ kn
ln a −
∑ cos nθ
2πε 0
2πε 0 n =1 n
where Schaum’s Eq. 23.22 was used in the last step to find the required Fourier series, so that
1
λ
a0 ≡ C0 + D0 ln R =
ln a,
2
2πε 0
=
bn ≡ An R− n = 0, and
λ 1  R n
.
2πε 0 n  a 
The fact that a pure cosine series was obtained makes sense in light of the symmetry about the xaxis in this problem. There remains a degree of freedom about how the value of a0/2 is split
between C0 and D0 . A particularly nice choice is to require ψ to be finite at infinity. Noting that
λ
ψτ → −
ln r and ψ S → D0 ln r as r → ∞ ,
2πε 0
we see that we can “cancel” these infinities by putting
λ
λ
⇒ C0 =
D0 =
(ln a − ln R).
2πε 0
2πε 0
Thus
2
λ
λ
λ ∞ k˜ n
a
˜ ≡ R <1
where
ψS =
ln +
ln r −
cos
θ
n
k
∑
2πε 0 R 2πε 0
2πε 0 n =1 n
ar
an ≡ Bn R− n = −
=
λ
λ
λ
a
ln +
ln r +
ln(1 − 2 k˜ cosθ + k˜ 2 ) using Eq. 23.22 again
2πε 0 R 2πε 0
4πε 0
R2
λ
λ
a
˜ cosθ + a˜ 2 ) where 㘠≡
ln +
ln(r 2 − 2 ar
a
2πε 0 R 4πε 0
and so the final solution is
λ
λ
a
λ
˜ cosθ + a˜ 2 )
ψ ( r,θ , z ) = −
ln(r 2 − 2 ar cosθ + a 2 ) +
ln +
ln(r 2 − 2 ar
4πε 0
2πε 0 R 4πε 0
for the potential everywhere outside the cylinder. We see that this solution is equivalent to an
image line charge of linear density –λ located at a distance ã < R from the origin along the xaxis inside the cylinder. The remaining term λ ln( a / R) / 2πε 0 is then the potential at infinity. By
varying this constant, we can in fact adjust the potential on the conducting cylinder to be any
other constant value we wish instead of zero.
=
Problems
(i) The image method begins by assuming a solution of the form
λ
λ
˜ cosθ + a˜ 2 ) + c
ψ (r,θ , z ) = −
ln(r 2 − 2 ar cosθ + a 2 ) +
ln(r 2 − 2 ar
4πε 0
4πε 0
where ã and c are unknown constants. Determine their values by solving the two equations
ψ ( R, 0, z ) = 0 and ψ ( R, π , z ) = 0
for these two unknowns, to verify the above solution.
(ii) Now explicitly verify that ψ ( R,θ , z ) = 0 for all θ, not just the two values chosen in Problem
(i). This is a matter of simple algebraic manipulation.
(iii) Find the surface charge density σ(θ) on the cylinder by taking the negative gradient of
ψ (r,θ , z ) to get the electric field, evaluating this field at the surface of the cylinder, and
multiplying the radial component thereof by ε0, as you learned in E&M I. Hint: The field at the
surface of the cylinder has to turn out to be purely radial because the conductor is an
equipotential. Use this as a check on your math!
(iv) For extra credit, multiply this surface charge density by Rdθ and integrate (using tables)
around the circle to verify that you get the expected image linear charge density, –λ .