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I have thought over this problem but I haven't found the solution: There is an electric charge $q$ at a distance $d$ from a conducting slab with thickness $t$, the problem is to find the potential everywhere, assumming that the slab is grounded. So, on the side of the slab where the charge is I think one may use the image method solution to find the potential in this region of space. Inside the slab $\phi =0$, but the question remains how to find the potential on the other side? The method of images does not seem to work on that side(?), and other solutions seem difficult to implement. Charge conservation seems to imply that the total charge induced on each surface of the slab is $\pm q$. Is there any easy or semi-easy way to do it? I guess my question would be: what is the most straightforward way to do it? Any help is greatly appreciated.

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up vote 1 down vote accepted

Let the "other side" exist in the lower-half space $z<0$. The conductor imposes the following boundary condition on the potential $\Phi = \Phi(x,y,z)$: \begin{align} \Phi(x,y,0) = 0. \end{align} The charge density vanishes for $z<0$, so we simply need to solve the Laplace equation $\nabla^2\Phi = 0$ for the region $z<0$ with this boundary condition at $z=0$. Note that $\Phi = 0$ is a solution to this equation satisfying the desired boundary condition, so this is the solution by uniqueness; \begin{align} \Phi(x,y,z) = 0, \qquad \text{for all $z<0$}. \end{align}

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Considering a spherical conducting shell of radius $R$ and thickness $t$, with a charge $q$ at a distance $d$ outside. So, what is the potential inside the shell? Extending $R\to\infty$ now and you get it.

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I am afraid this will not work. The electric field inside a conducting spherical shell will always be zero (conductors shield electric fields), hence the surface charge densities in the inner shell will be zero. This cannot be the case for the infinite slab, so the solution cannot be obtained in this way. – Rogelio Molina Sep 6 '13 at 9:55
I stand corrected. Indeed the method you gave works well as the charge density on the opposite side to the plate will vanish, as joshphysics pointed out. It is remarkable that the thickness plays no role whatsoever. – Rogelio Molina Sep 6 '13 at 12:28

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