I know how to solve Laplace's equation for a point charge in front of a grounded conducting infinite plane. But I want to know what happens (both physics and math) when the infinite conducting plane isn't grounded, or is connected to a potential $V$.
[Physics] Method of image charges for a point charge and a non-grounded conducting plane
chargeelectrostaticsmethod-of-imagespotential
Best Answer
The problem is addressed in Electrodynamics Fulvio Melia Example 2.1 page 39, using Green function techniques. However, I cannot fully understand his final solution. Let us define the plane held at a constant potential $V_0$ to be the $z$-$y$ plane. The charge $q$ is at the distance $a$, namely at $(a,0,0)$. The potential at large distances is zero. The solution (Eq. 2.54) is
$$V(x,y,z) = V_{image} (grounded) + V_0 f(x,y,z).$$
The first term is the usual image method solution, for the case where the conducting plane is grounded, namely
$$V_{image}(grounded) = q\left[\left[(x-a)^2 + y^2 + z^2\right]^{-1/2} - \left[(x+a)^2 + y^2 + z^2\right]^{-1/2}\right],$$
which gives zero when $x=0$. In other words this is the solution of the Poisson equation, but the boundary conditions are not those required.
The second term, $V_0 f(x,y,z)$ is the solution of Laplace equation, and $f(x,y,z)$ is $V_0$ independent.
According to the text book Eq. (2.54) there
$$f(x,y,z) = (x/2 \pi) \int dz' \int dy' \left[x^2 + (y-y')^2 + (z-z')^2\right]^{-3/2}.$$
The integrals over the primed variables stretch from minus infinity to plus infinity and this function is V_0, q and the length scale a independent.
The solution is mysterious for me. Namely, we can get rid of $y$ and $z$ in the integral (for any finite $y$ and $z$) by change of variables of integration. Then the integral can be solved (for $x>0$) and the solution is $f(x,y,z)= x/ x = 1$. Well, that is wrong (or subtle) since that would mean that we only need to add to the usual image solution a constant $V_0$. This does not satisfy the boundary condition at large distances from the plane where the potential is zero.
Maybe the integral for the dimensionless function f(x,y,z) should be regularised somehow.
Since the image part of the solution, $V_{image}(grounded)$, takes care of the charge in the Poisson equation, we can find a solution that is a linear combination of $V_{image}(grounded)$ and some other function, the latter solves the Laplace equation. Then let the boundary at $x=L$ be grounded and we want to take $L\to \infty$ and on x=0 we have V_0 (this seems like a legitimate trick). The Laplace equation for this problem gives
$$V(x,y,z) = -V_0 x/L + V_0,$$ and if we take $L\to \infty$ we get the result just mentioned (for any finite $x$) namely a constant (which does not satisfy the boundary condition at infinity). So I think the limit of a large system ($L\to \infty$) is creating a certain non-trivial problem, which ultimately is related to the Green function technique used by Fulvio Melia. Any ideas?
Of-course if f(x,y,z) is unity for any finite distance x, this would mean that switching from grounded plane to non-grounded one (compared to zero potential at infinity) has practically no effect. This hypothesis was postulated in some of the previous discussions.