The electrostatic energy of a point charge
interacting with a neutral metallic hemisphere
Michael Levin1 and Steven G. Johnson2
1
2
Department of Physics, University of Maryland, College Park, MD 20742
Department of Mathematics, Massachusetts Institute of Technology, Cambridge MA 02139
(Dated: April 20, 2011)
In this note, we review the calculation of the electrostatic interaction between a point charge
and a neutral conducting hemisphere, which is related to a classic problem in electrostatics first
studied by Lord Kelvin. This interaction exhibits an interesting phenomena that was only pointed
out recently, however: the charge is actually repelled from the hemisphere for a certain range of
separations, unlike the attractive interaction between charges and most neutral metallic surfaces
(e.g. spheres or planes).
2
I.
INTRODUCTION
In a recent article [1], we considered a simple question in electrostatics—can a point charge be repelled from a
neutral metallic conductor of any shape? Intuitively, the answer might seem to be no: a positive charge next to a
neutral conductor induces negative charges on nearby surfaces and positive charges on farther surfaces, and hence
would seem necessarily attracted. Surprisingly, however, repulsion can occur for a point charge next to certain nonconvex shapes, even when the charge and the conductor are on opposite sides of a separating plane (so that there is
no ambiguity about the direction of the force).
In particular, we developed a set of repulsive examples by a simple analytical argument [1]: a thin metallic surface
placed on an equipotential surface of the charge does not interact with the charge, and hence when the charge is moved
away to infinity there must be a combination of repulsive and attractive forces at different separations so that the net
work is zero. With a simple example, we also illustrated the flaw in the intuitive argument about negative charges
being induced on nearby surfaces—while this is true, these surfaces can be arranged to exert forces at oblique angles
so that their net attraction is outweighed by the repulsive effect of positive charges at farther surfaces. Finally, we also
presented an illustrative 2d example system in which the repulsive effect can be demonstrated by explicit analytical
solutions for the charge at any separation, using standard complex-analysis methods. In this note, we supplement
that illustration by a second example that can also be solved analytically, that of a charge interacting with a thin
metal hemisphere.
Such electrostatics problems, in themselves, are nothing new, but we are not aware of any previous author who
commented on the existence of repulsive solutions. In particular, the problem of the charge distribution on a hemispherical conductor was famously solved by Lord Kelvin in 1848 [2, 3], and Kelvin’s result was later reproduced in
part by Maxwell’s well-known textbook [4, pp. 221–225]. Jeans [5, p. 289] points out that Kelvin’s solution leads to
a beautifully simple expression for the capacitance C of a hemisphere of radius R: C = 4πε0 R(1/2 + 1/π) (below,
we employ cgs units so that 4πε0 is replaced by 1). Jeans and Maxwell focus on Kelvin’s results for the isolated
hemisphere and for grounded hemispheres in the presence of a point charge located on the coincident sphere, but
Kelvin also presents a solution for the charge density induced on a grounded hemisphere by a point charge at any location. However, Kelvin only evaluated his formulas numerically for the limit of a flat grounded disk. Moreover, direct
application of Kelvin’s solution to the case of a neutral (rather than grounded) hemisphere requires a cumbersome
integral to be evaluated in order to determine and subtract the net induced charge. Therefore, we instead adapt parts
of Kelvin’s basic approach—evaluating the case of a disk (or its inverse, a circular hole in a metallic plate) and then
transforming to a hemisphere via inversion—to slightly more modern techniques in order to simplify the calculations.
II.
A POINT CHARGE INTERACTING WITH A PLATE WITH A HOLE
Our strategy for solving the metallic hemisphere problem is to first analyze a simpler, but closely related, geometry
and then map this geometry onto the hemisphere problem using a 3d inversion transformation.
This simpler, but closely related, geometry consists of a point charge interacting with a metallic plate with a hole.
We assume that the point charge q lies on the z axis at some height z0 > 0 while the metallic plate is positioned in
the z = 0 plane with a circular hole of unit radius centered at the origin. We wish to find the electrostatic potential
created by the charge. Working in cylindrical coordinates (ρ, z, θ), this potential can be written as Φz0 (ρ, z).
Following Jackson [6, pp. 129–135], we write Φz0 = Φ0z0 + Φ1z0 , where Φ0z0 is the potential created by a charge
interacting with an unpunctured metallic plate, and Φ1z0 represents the additional contribution coming from the hole.
Clearly, Φ0z0 can be obtained using image charges:
!
q
q
Φ0z0 (ρ, z) = p
−p
Θ(z)
(1)
ρ2 + (z − z0 )2
ρ2 + (z + z0 )2
As for Φ1z0 , this function satisfies Laplace’s equation
∂ 2 Φ1z0
∂Φ1z0
1 ∂Φ1z0
+
=0
+
∂ρ2
ρ ∂ρ
∂z 2
(2)
in the upper half space z > 0, with the mixed boundary conditions
1
∂z Φ1z0 (ρ, 0) = − ∂z Φ0z0 (ρ, 0+ ) ; 0 ≤ ρ < 1
2
1
Φz0 (ρ, 0) = 0 ; 1 < ρ < ∞
(3)
3
The question reduces to solving this mixed boundary value problem. Luckily, the general solution for this kind of
problem is already known. Using a result reviewed by Duffy [7, pp. 268–270], the solution is given by:
Z ∞
1
dkA(k)e−k|z| J0 (kρ)
(4)
Φz0 (ρ, z) =
0
where
A(k) =
1
Z
dt sin(kt)x(t)
(5)
0
and
x(t) =
t
Z
t
2qz0
ρ
dρ p
2
t2 − ρ2 (z0 + ρ2 )3/2
dρ
0
Substituting in the expression for Φ0z0 (1) gives
x(t) =
Making the change of variables ρ =
1
π
Z
ρ∂z Φ0z0 (ρ, 0+ )
p
t2 − ρ 2
1
π
0
(6)
(7)
p
t2 − (z02 + t2 ) sin(θ)2 , the integral can be easily evaluated:
x(t) =
2qt
π(z02 + t2 )
(8)
Substituting this into (4-5) gives
Φ1z0 (ρ, z) =
Z
∞
dke−k|z| J0 (kρ)
0
Z
1
dt sin(kt)
0
2qt
+ t2 )
(9)
π(z02
In general, this integral is hard to evaluate. However, along the z-axis where ρ = 0 (the case we are primarily
interested in), it can be evaluated in closed form. Evaluating the k integral first, we find
Z 1
t
2qt
dt
(10)
Φ1z0 (0, z) =
2 + t2 ) z 2 + t2
π(z
0
0
2q
(z arctan(1/z) − z0 arctan(1/z0 ))
(11)
=
2
π(z − z02 )
We conclude that
Φz0 (0, z) =
Φ0z0 (0, z)
+
Φ1z0 (0, z)
=
q
q
−
|z − z0 | |z + z0 |
Θ(z) +
2q
g(z, z0 )
π
(12)
where
g(u, v) =
u arctan(1/u) − v arctan(1/v)
u2 − v 2
(13)
This is the potential when z0 > 0; when z0 < 0, one obtains the same formula, but with the replacement Θ(z) →
Θ(−z). The two cases can be succinctly summarized by the general result:
q
2q
q
Φz0 (0, z) =
Θ(zz0 ) + g(z, z0)
−
(14)
|z − z0 | |z + z0 |
π
III.
MAPPING TO THE HEMISPHERE PROBLEM
We now map the problem of a charge interacting with plate with a hole onto the problem of a charge interacting
with a hemisphere. We accomplish this via the map
h(x) =
2(x + ẑ)
− ẑ
|x + ẑ|2
(15)
4
where ẑ is the unit vector in the z direction.
This map describes a 3d inversion transformation composed with a translation in the z direction. It is easy to check
that h maps the hemisphere of unit radius centered at 0, positioned in the lower half space z < 0, onto the z = 0
plane with a circular hole of unit radius, centered at the origin. Therefore, if Φ is the potential created by a charge
at position h(y ẑ) interacting with a plate with a hole, then by the standard 3d inversion method (i.e. the Kelvin
transform), the transformed potential
Φhemi = Φ(h(x)) ·
2
|x + ẑ|
(16)
is the potential created by a charge at position y ẑ interacting with a charged hemisphere.
This isn’t quite what we want, since we are interested in the case of a charge interacting with a neutral hemisphere.
To solve the neutral hemisphere problem, we will let Φ be the potential created by two charges—a charge q1 at h(y ẑ)
and a charge q2 at h(∞) = −ẑ. Then, by choosing q2 appropriately, we will tune the potential on the hemisphere to
0.
The first step is to write down the potential Φ created by a charge q1 at h(y ẑ) and a charge q2 at h(∞) = −ẑ.
Focusing on the case where x lies on the z-axis, x = (0, 0, z), we can apply the result (14) to obtain
q1
q2
q1
q2
Φ(z) =
Θ(zh(y)) +
Θ(−z)
−
−
|z − h(y)| |z + h(y)|
|z + 1| |z − 1|
2q1
2q2
+
g(z, h(y)) +
g(z, 1)
(17)
π
π
where h is the map h restricted to the z-axis:
h(y) =
2
1−y
−1=
y+1
1+y
Substituting this into (16) gives a hemisphere potential
2Θ(h(z)h(y))
2Θ(−h(z))
q1
q2
q2
q1
hemi
−
+
−
Φ
(z) =
|h(z) − h(y)| |h(z) + h(y)|
|z + 1|
|h(z) + 1| |h(z) − 1|
|z + 1|
4q1 · g(h(z), h(y)) 4q2 · g(h(z), 1)
+
+
π|z + 1|
π|z + 1|
Expanding the first four terms gives
1
q1 |y + 1| q1 |y + 1|
Φhemi (z) =
Θ(h(z)h(y)) + q2 1 −
Θ(−h(z))
−
|z − y|
|yz − 1|
|z|
4q1 · g(h(z), h(y)) 4q2 · g(h(z), 1)
+
+
π|z + 1|
π|z + 1|
(18)
(19)
(20)
Now all that remains is to choose q1 , q2 appropriately. We choose q1 = q/|y + 1| so that the charge at position y has
magnitude q. The result is:
1
q
q
hemi
Φ
(z) =
Θ(h(z)h(y)) + q2 1 −
Θ(−h(z))
−
|z − y| |yz − 1|
|z|
4q · g(h(z), h(y)) 4q2 · g(h(z), 1)
+
+
(21)
π|y + 1||z + 1|
π|z + 1|
As for q2 , this needs to be chosen so that the hemisphere is neutral. To this end, we compute the asymptotic behavior
of Φhemi in the limit z → ∞. A little algebra shows that
4q · g(h(y), 1)
q
1
hemi
Θ(−h(y)) − q2 +
·
q−
+ q2 (1/2 − 1/π)
(22)
Φ
∼ q2 +
|z|
|y|
π|y + 1|
as z → ∞. The constant term indicates that we need to shift our potential Φhemi → Φhemi − q2 so that it vanishes at
∞. As for the 1/|z| term, this should have a coefficient of q if the hemisphere is neutral. Therefore we need
q
Θ(−h(y)) + 4q·g(h(y),1)
−q
q − |y|
π|y+1|
(23)
q2 =
1/2 + 1/π
5
Making these substitutions, we derive the z-axis electrostatic potential of a charge q at position y ẑ interacting with
a neutral metallic hemisphere of unit radius:
4q · g(h(z), h(y))
q
q
Θ(h(z)h(y)) +
Φhemi
(z)
=
−
y
|z − y| |yz − 1|
π|y + 1||z + 1|
q · f (y)f (z)
(24)
+
1/2 + 1/π
where
4g(h(z), 1)
1
Θ(−h(z)) +
−1
f (z) = 1 −
|z|
π|z + 1|
IV.
(25)
COMPUTING THE ELECTROSTATIC ENERGY
Our goal is to find the electrostatic energy U hemi (z) of the charge interacting with the hemisphere. This can be
easily obtained from the electrostatic potential (24) using the relation
q
q
hemi
hemi
U
(z) =
Φy (z) −
(26)
2
|z − y| y=z
Substituting in the explicit form of the potential (24) we find the following expression for the electrostatic energy of
a charge q at position z interacting with a neutral hemisphere of unit radius:
1
q2
h(z)
q 2 · f (z)2
q2
+
(27)
arctan
−
+
U hemi(z) = − 2
|2z − 2| π(1 − z 2 )
h(z)
h(z)2 + 1
1 + 2/π
[The functions f, g, h are defined in equations (25),(13),(18).] A plot of U hemi(z) is shown in Fig. 1. We can see
from the figure that the force is attractive for large z and repulsive for small positive z, with a sign change at
z ≈ 0.629083067265R. The force changes sign again at z = 0, before diverging to −∞ as z → −R. We can also see
that U hemi (0) = 0. These properties are in complete agreement with the general results and arguments given in our
paper [1].
V.
COMPUTING THE CAPACITANCE OF THE HEMISPHERE
As an aside, we compute the capacitance of the hemisphere, for comparison with Jeans [5]. To do this, we consider
the potential (21) in the case q = 0: in that case, Φhemi is the potential of a charged hemisphere in vacuum.
To extract the capacitance, we examine the asymptotic behavior of Φhemi(z) in the limit z → ∞. This can be
accomplished by setting q = 0 in (22):
Φhemi ∼ q2 −
q2
· (1/2 + 1/π)
|z|
(28)
Examining the 1/z term, we see that Φhemi describes a hemisphere with charge
Q = −q2 (1/2 + 1/π)
(29)
On the other hand, the constant term tells us that the voltage of the hemisphere is
V = Φhemi (z = −1) − Φhemi(z = ∞) = −Φhemi (z = ∞) = −q2
(30)
C = Q/V = 1/2 + 1/π
(31)
Taking the ratio, we find
in agreement with Jeans’ answer.
6
0
Uhemi(z) (units of q2/r)
−0.002
−0.004
−0.006
−0.008
−0.01
−1
0
1
2
3
4
z/R
FIG. 1. Exact electrostatic interaction energy U hemi (z) of a charge interacting with an infinitesimally thin neutral metallic
hemisphere of radius R (27). The hemisphere is centered at (0, 0, 0) and lies within the lower half space z ≤ 0, while the charge
is at position (0, 0, z) with z ≥ 0.
ACKNOWLEDGEMENTS
We are grateful to Dr. M. T. Homer Reid at MIT for inspiring us with a related calculation of an electrostatic
dipole above a plate with a hole, and to Prof. Wayne Saslow at Texas A&M for helpful discussions of the work of
Maxwell and Thomson (Lord Kelvin).
[1] M. Levin and S. G. Johnson, “Is the electrostatic force between a point charge and a neutral metallic object always
attractive?,” arXiv.org e-Print archive, p. arXiv:1007.2175, 2010.
[2] B. W. T. Kelvin, “Determination of the distribution of electricity on a circular segment of plane or spherical conducting
surface, under any given influence,” in Reprint of Papers on Electrostatics and Magnetism, ch. XV, pp. 178–191, Macmillan,
1872.
[3] A. Russell, “The eighth Kelvin Lecture: Some aspects of Lord Kelvin’s life and work,” J. Institution of Electrical Engineers,
vol. 55, no. 261, pp. 1–17, 1916.
[4] J. C. Maxwell, A Treatise on Electricity and Magnetism. Oxford: Clarendon Press, 1873.
[5] J. H. Jeans, The Mathematical Theory of Electricity and Magnetism. Cambridge Univ. Press, 1908.
[6] J. D. Jackson, Classical Electrodynamics. New York: Wiley, third ed., 1998.
[7] D. G. Duffy, Mixed Boundary Value Problems. Applied Mathematics and Nonlinear Science, Boca Raton, FL: Chapman &
Hall/CRC, 2008.