A unit charge is placed outside a grounded conducting unit radius circle at the origin. What is the field outside the circle?

This is not the obvious problem you immediately think it is.

A classic, perhaps the classic electrostatic induction example is determining the field around, and charge on, a grounded conductive sphere induced by a point charge. The example goes all the way back to Lord Kelvin, and it's a staple in every electromagnetic text.

The boundary conditions are simple. First, the potential is constant on the sphere's surface, because it's a conductor. It's grounded so infinity has the same potential as well (usually taken to be 0). If the inducing point charge is outside the sphere, then Poisson's equation holds throughout the region outside the sphere with a delta function at the charge location. If the inducing charge is inside the sphere, the same Poisson equation holds inside the sphere instead.

The answer is formed formed by placing a virtual charge at the spherical inverse point to the inducing charge and setting its magnitude to satisfy the boundary conditions. The potential is written as the sum of the point potentials of both the original and inducing charge. The induced charge density is given by $\partial V\over \partial r$ evaluated at the unit $r=1$.

If the problem is of a unit charge inside the sphere at point (a, 0, 0) then the answer (for the field inside the sphere) is given by adding a virtual charge at (1/a, 0 0 ) with charge of -a.

If the problem is of a unit charge outside the sphere, the answer for the field outside the sphere is given by adding a virtual charge at (1/a, 0, 0) with magnitude -1/a.

Classic examples, classic homework. Let's move on.

In 2D, image charges also work for the case of the inducing charge inside the conductive circle. The virtual charge is at (1/a, 0) and has magnitude 1. This is just like 3D except for the magnitude. Even though in 2D potential follows a logarithmic falloff, the equipotentials are still circles. Here we compute a constant positive net potential on the circle, but we allow ground (located at infinity) to have the same potential so our boundary conditions are fufilled and the problem is solved.

Now we get to the tricky question: what happens with an inducing charge outside of the circle at (a, 0)? We proceed, adding a virtual charge inside the circle at (1/a, 0). Its magnitude is -1. This creates a constant potential on the circle as desired.

So is that our answer? Poisson's equation is fulfilled outside the circle. Check! Conductive circle is an equipotential, check! Circle is grounded, so it's the same potential as infinity. Uhoh. It's not. Infinity has a potential of 0 (in far field the +1 and -1 charges cancel out). But our circle is not at potential 0. It's lower.
Placing a second charge at the circle's center to "correct" the boundary condition does not work.. the total of the virtual charges must still be -1, and the original induced virtual charge must be -1 to make an equipotential, so any virtual charge at the origin must have magnitude 0.

This puzzle has been stumping me for days. A dozen online pages gleefully explain the problem in 3D (both from inside and outside the sphere) and a couple explain the 2D problem from the inside. None from the outside. Not a single text discusses this problem! Not Griffiths, Jackson, Stratton, Schwartz, or Eyges. Smythe's classic text has a good example of an ungrounded circle conductor with outside inducing charge, which he solves as a limit of a dielectric as its constant goes to infinity, but that's not the same as this problem as a grounded circle.

What is the solution to this problem? It's a real puzzle and surprising that nobody has discussed it even in physics texts.


3 Answers 3


Were the circle held at fixed potential $V$, the solution for the image charge of a cylindrical conductor would work (See Jackson 2.11, with $B\rightarrow 1$ and $\tau \rightarrow q$, but $4\pi$ stays the same!). This line of reasoning gives us $q'=-q$ and $r' = \frac{1}{a}$, just as you have found.

However, I think that holding the circle at $V=0$ changes the boundary condition $\Phi(r\rightarrow \infty)$ to a nonzero value, since the potential due to the single test charge is non-vanishing at infinity (tricky logarithms!). As a result, $q'\neq q$ and the position of the image $r'$ doesn't have a clean analytical solution.

  • $\begingroup$ And therein lies the puzzle. That non-vanishing potential at infinity means the positive and negative charges in the whole system must sum to 0. But the canonical image charge must have a magnitude of -1.. leaving us with no flexibility. Yet there is a solution since the Poisson equation is well defined and has simple and consistent boundary conditions. $\endgroup$
    – Max Filoli
    Commented Dec 15, 2016 at 6:37
  • $\begingroup$ The Jackson problem is a great example! Like Smythe, it discusses the simple UNGROUNDED conductor. Grounding it causes the puzzling conundrum. $\endgroup$
    – Max Filoli
    Commented Dec 15, 2016 at 6:43
  • $\begingroup$ Since potential=0 at infinity does imply q'=-q, it seems paradoxical that a non-vanishing potential implies q'=-q also; on the contrary, I expect V(infinity) to have a functional form ~ln(r) as r->infinity. I tried to show this using the methods of the Jackson solution but imposing a boundary condition V(r->infinity) = f(r). This allowed q'=-q but also a trivial solution V = log(r)/(2*pi*eps) - admittedly i do not know where to go from there... I will keep trying and update with progress later. $\endgroup$
    – forky40
    Commented Dec 15, 2016 at 21:14
  • $\begingroup$ Thanks, forky40! The mystery is deep. My latest attempt tried replacing image charges with a general multipole expansion.. but I can't satisfy both boundary conditions. Which makes me start thinking there is no answer, the setup is physically impossible? But the problem itself seems well defined and not extraordinary. $\endgroup$
    – Max Filoli
    Commented Dec 16, 2016 at 6:07

I am posting my own answer here, hopefully to gather suggestions of whether this is an actual answer.

A pair of charges with magnitude -1 at (1/a, 0) and +1 at (a,0) makes an equipotential on the origin centered circle R with potential proportional to log(R/a). At infinity, the charges cancel out a produces a potential of 0. This difference is causing our problem.

We solve it by adding a second pair of charges of magnitude m at (1/b, 0) and -m at (b, 0). These create their own equipotential proportional to -m log(1/b). We choose m equal to log(1/a)/log(1/b) = log(a)/log(b) so that the net potential on the circle is 0, and the potential at infinity is also zero.

But this violates the restructions that image charges cannot be inside the domain we're evaluating. But we can choose b, so this can be "hidden" by taking the limit as b goes to infinity.

Thus the boundary conditions are met. Potential on the circle is constant 0, matching potential at infinity. Poisson's equation holds in the finite region outside the circle, with the required singularity at our inducing charge. The problem is solved!

The unsatisfying conclusion, however, is that as our second two images charges reach the limit as b goes to infinity, their magnitude goes to 0. So we don't need the third and fourth image charges at all. I'm not even sure what this means in consequence since if the extra image charges are not included, our circle isn't grounded. Adding the infinitesimal magnitude image charge pair seems to allow us to arbitrarily reset the circle's potential to whatever we want.


Not all puzzles have a solution. Your hidden assumption is that there always should be a solution with the object's surface at the same potential as infinity, a situation we then define as "grounded". (We could also define grounded as just requiring $V=0$ at the surface, but let's stick with the first definition for the moment.)

We then have to face the fact is that there is no need for such a solution to exist. Electromagnetism does not become inconsistent if it doesn't exist. We know that there actually will be such a solution if the number of dimensions $D$ is greater than $2$, but not necessarily for $D=2$. And $D=2$ is not special, for $D=1$ it is the same.

Let's look at the separate logical steps.
A) We want to solve Poisson's equation for a charge distribution on some surface with a given pre-existing point charge which is not on that surface.
B) We require the resulting potential to be constant on the surface, since it is conducting. This leaves two things to fix:
1: The total charge on the surface. There is always a new solution if we add some amount of charge and let it settle.
2: An arbitrary harmonic function (solution of $\nabla^2 V=0$) can be added to the potential, without changing anything in the charge distribution.
C) A straightforward way to fix things would be to specify the total charge $Q_\text{surf}$ on the surface, and to specify how $V$ behaves on some enclosing boundary chosen outside the whole system. Uniqueness can be proven for those conditions.
D) But we want to simplify the boundary condition by only requiring $V\rightarrow V_\text{inf}$ if $r\rightarrow\infty$.

Clearly, step D) in the procedure is where things go wrong: this is not possible for $D=1$ or $2$, where the potential will not approach any finite value $V_\text{inf}$ unless we only consider solutions with a net total charge $0$. Again, it should be stressed that this is no inconsistency (I'd say it is not even a paradox), because there is no need that this simplification should be possible. It is merely a mathematical technicality of how exactly to specify the boundary conditions. For $D=2$ we could simply use instead: $$ \begin{align} V& \underset{r\rightarrow\infty}{\longrightarrow}\ C_\text{inf} \cdot \ln r + O({\small \frac1 r}) \end{align}$$ That would rule out all of the (infinitely many) harmonic functions we could otherwise add to the potential, like $x,y, y^2\!-\!x^2$, $xy$, etc.

So what does it mean? Just that we cannot use the concept "grounded surface" defined as "at the same potential as infinity" to fix the freedom in finding the solution. We can find the solution and fix all its degrees of freedom, so there is no real problem, but we might still ask if there is another definition of grounded so that we could still use the concept. That's possible if we choose some radius $R_\text{ref}$ and define grounded as:
$\qquad$ "the same potential as the average over a circle of radius $R_\text{ref}$".
This does not require a big conductive cylinder to be physically present at radius $R$ (which would change the solutions!) since the integral over the circle will vanish for all multipole functions other than $\ln r$. So we would only change the "grounded" definition $V_\text{surf}=V_\text{inf}$ into $V_\text{surf}=C_\text{inf}\ln R_\text{ref}$


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