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.