Electric induction in a non-grounded plane conductor

Question

Supose we have a charge $+q$ which is held at a distance $d$ from the plane $z=0$ not grounded. Very similar to the classic case: Electric induction in a grounded plane conductor

But the infinite plane is not grounded. Which would be the charge distribution in the plane then?

I assume that the charge conservation will give it null total charge. But unlike the sphere, which is a closed surface, we can't just add $\sigma = + \frac qA$ because that would just be zero.

How can we solve Laplace's equation for this case? Should we solve it for a finite plane and take the limit?

EDIT: it seems to me we only need to solve Neuman's Laplace equation for $V(z=0) = V$ and then impose the total charge in the plane to be zero.

Answer 1

To find the surface charge you can try first to find the potential in the region $\varOmega=\{\mathbf{r}\in \mathbb{R}^3: z>0\}$, where you have the point charge, using the Green function. Remember:

$$V(\mathbf{r})=\int_\varOmega G(\mathbf{r},\mathbf{r}^\prime)\frac{\rho(\mathbf{r^\prime})}{\varepsilon_0}dV'-\int_{\partial\varOmega} V_0\nabla G(\mathbf{r},\mathbf{r}^\prime)\cdot\mathbf{n}'dS'$$

Then apply that the surface density causes a discontinuity in the electric field. In $\mathbf{E}(z=0^-)=0$ and $\mathbf{E}(z=0^+)=-[\nabla V]|_{\displaystyle z=0}\quad$, so that $\sigma=\varepsilon_0\mathbf{e_z}\cdot\left[\mathbf{E}(z=0^+)-\mathbf{E}(z=0^-)\right]$