Problem 3.16 from "Introduction to Electrodynamics" by D. J. Griffiths.
Five sides of a cube are at zero potential. One remaining side (insulated from others) is at potential $V_0$. What is the potential at the center of the cube?
I know how to find analytical solution for this problem (by solving Laplace equation for the given boundary conditions). Is there any other (easier/conceptual) way to find the potential at the "center"? The numerical value comes out to be very close to $V_0/6$. Will it be exactly $V_0/6$? If yes, then what is the reasoning behind it?
Best Answer
It seems that the answer should actually be exact. The proof goes as follows.
Let's define charge distribution $C_1$ the distribution the system (the whole system, not just one side) would have, if boundary conditions are satisfied and 1st side of the cube has potential $V_0$. Let's define similarly $C_2$ ... $C_6$. Now these distributions have an obvious (but very useful) property. Because of superposition, if the system had before some charge distribution and thus some point, which lies on for example surface 1, had previously potential $\phi_1$, after adding $C_i$, it would have potential $\phi_1 + V_0$ if $i = 1$ and $\phi_1$ if $i \neq 1$ (directly from definition of $C_1$).
Because of symmetry, adding $C_i$ to the system for any $i$, must increase the potential in the center by a constant amount $\Delta V$. However, because of the "obvious" property, if the cube with initially no charges is charged with distribution $C_1 + ... + C_6$, the boundary of the resulting cube is uniformly at potential $V_0$ - which implies that at the center it is also $V_0$ (it is effectively in a metal cage). Thus $\Delta V$ must have been $V_0 / 6$, which is the answer.