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.

  • $\begingroup$ Your assumption that the plane should have null total charge is not right. A perfect conductor is able to redistribute its electrons so that the electric field vanishes inside and can have as much net charge as necessary to obtain equilibrium. $\endgroup$
    – Urb
    Commented May 19, 2020 at 11:48
  • $\begingroup$ I disagree. For me you are describing the grounded case where the charges can come from Earth. If the plane had no charge in the beggining it should have no net charge in the end. This is how we solve the non-grounded sphere. $\endgroup$ Commented May 20, 2020 at 10:07

1 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 $\sigma=\varepsilon_0\mathbf{e_z}\cdot\left[\mathbf{E}(z=0^+)-\mathbf{E}(z=0^-)\right]$

  • $\begingroup$ ok I will try to solve the problem directly as you described! $\endgroup$ Commented May 20, 2020 at 10:08

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