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Imagine an infinitely long conducting "trough," as shown in the figure. The two sides are grounded, and the bottom strip is maintained at potential $V_0$. Suppose we want to know the electric potential everywhere between the plates. The electric potential is a solution to Laplace's equation, which also satisfies the required boundary conditions. The boundary conditions are $V(x,y,z) = 0$ on the left and right plates, $V(x,y,z)=V_0$ on the bottom plate, and $V(x,y,z)\to 0$ at $\infty$.

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My question is about this last boundary condition. On one hand it seems reasonable, and I've been using it for years. However, I was recently asked why $V$ goes to zero at infinity, and I couldn't answer the question satisfactorily. The issue is that the bottom plate seems to resemble an infinite line of charge when we are far away from it, but we know there are problems with setting $V=0$ at infinity when we're dealing with infinite charge distributions. In the case of a line of charge, $E\propto 1/r$ where $r$ is the distance from the line of charge. If we integrate $E$ from $\infty$ to some finite $r=a$, we find $V(a)=\infty$. (I.e. $V = -\int_\infty^a E dr$ means our calculation of $V$ involves something with $\ln(\infty) - \ln(a)$, which is infinite.) With this in mind, in the case of the conducting plate in the figure, it seems that if we set $V=V_0$ at the bottom plate, then $V$ should decrease forever as we get farther away. That might lead one to say that $V\to -\infty$ as we get infinitely far from the bottom plate.

So, why does $V\to 0$ as we get infinitely far away from the bottom plate? If we had an infinite line of charge, the potential would decrease without bound as we moved away from the line charge. The difference in the case of the conducting sheets must be related to the sides being grounded (thus kept at $V=0$)?

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If you solve Laplace equation outside the trough, you will see that a large part of the electric field lines emerging from the ground plane will end on the sidewalls. Therefore, as you say, the potential goes to zero faster than in the case of a line charge. You would have a similar case with two closely spaced line charges of opposite sign.

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