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I am looking for some help with a problem of Electrostatics.

Let us consider three parallel conducting slabs $P_1, P_2, P_3$ of a certain surface $S$, small thickness and mutual distance $d$. We know that the potential of the central one ($P_2$) is $\varphi_0 >0$, while the other two are connected to ground, i.e. their potential is $\varphi = 0$. The problem asks to determine the charge on all surfaces, neglecting any edge effects (i.e. treating the planes as infinite).

Using Gauss's Law and the fact that in a conductor the electric field is zero at equilibrium, it's easy to prove that facing charges are equal but with opposite sign (e.g. $Q_{12} = - Q_{21}$, with $Q_{12}$ the charge on $P_1$'s surface in front of $P_2$). Then, knowing the potential difference and again Gauss's Law (or evaluating the capacity of each capacitor) it is quite easy to determine the values of those charges.

The problem is I cannot find any mathematical relations proving that the "external" charges on $P_1$ and $P_3$ are zero, due to the fact that these conductors are grounded, which is what I "expect". The same problem does not appear for "finite" systems: for instance, if a charged sphere $Q$ is surrounded by a spherical shell connected to ground, we know for sure that the charge on the inner part of the shell is $-Q$, while the charge on the outer part is 0, since the potential of the shell is zero. In our case, the integral of the electric field cannot be performed, since we are neglecting edge effects. The only thing we can say is that, whatever these two charges will be, say $Q_{1ext}$ and $Q_{3ext}$, we will have $Q_{1ext}=Q_{3ext}$, since the electric field inside the conductor is again zero. In other words, it seems that there is no constraint on these two values, which can be even $\infty$.

How is that possible? What am I missing?

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First, we argue that the electric field outside the slabs is $0$. Assume a nonzero electric field $E$ exists just outside of $P_1$. It must be perpendicular to the slab everywhere, by symmetry. By Gauss's Law, it must be constant magnitude. But then, the potential of $P_1$ can't be $0$, which is a contradiction; hence, the electric field outside the slabs is $0$. (You can also show this from existence-uniqueness from the boundary conditions).

Simply apply Gauss's Law again to show $Q_{1\text{ext}}=Q_{3\text{ext}}=0$.

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  • $\begingroup$ "But then, the potential of P1 can't be 0." That is precisely what I do not understand. From an intuitive point of view, I guess there will be no charge, but it still does not seem obvious to me how to make that assertion. Of course if the slab was finite (as I mentioned for a sphere) and we had some symmetry, I could always evaluate the potential as an integral of the electric field up to infinity. $\endgroup$
    – Tailor
    Nov 7 '20 at 0:26
  • $\begingroup$ Another reasoning I could understand is the following, related to yours: with a constant magnitude electric field the potential difference between $P_1$ and a point at infinity would be $\infty$, and assuming the boundary at infinity is at zero potential (or, at least, at the same potential of $P_1$) we get $0 = \infty$, hence the contradiction. But again, this relies on the assumption of the potential at infinity. Besides, if $\varphi_{P1}=\varphi_{P3} \neq 0$, the same argument leads to infinities, so I guess it's a problem inherent to infinite systems. $\endgroup$
    – Tailor
    Nov 7 '20 at 0:37
  • $\begingroup$ @Tailor, an infinite slab has both translational and rotational symmetry, and just as in the case of a sphere, you can integrate the electric field along a suitably chosen path to determine the potential at any given point and/or take the limit as the distance goes to infinity. For a sphere, the best choice of path is radial, for a slab it is perpendicular. Otherwise there really isn't all that much difference between the two cases, you also have to assume that the potential at infinity is zero in order to solve for the sphere, right? $\endgroup$ Nov 9 '20 at 4:13

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