Potential at the center of a cubical box 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?
 A: 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.
A: For harmonic functions (solutions of the Laplace equation) there is a property called the mean value theorem.
It states that if you have a ball $B_R(x)$ of radius $R$ centered at $x$. The boundary of this ball is a sphere. Then the value of the harmonic function $\varphi$ at the center of the ball is given by a mean value of that function on the sphere:
$$\varphi(x)=\frac{1}{S_R}\int\limits_{\partial B_R(x)}\varphi\, dS $$
where $S_R$ is a surface of the sphere, in 3-dimensions it's $S_R=4\pi R^2$.
So you can expect that the value in the center of the cube will be approximately equal to mean value on its boundary. However you also may expect that there should be some corrections because the cube is not a ball. I suggest looking up the proof of the mean value theorem for possible insights. It basically is obtained from Green's identity,
$$\int\limits_V(\varphi\Delta\psi-\psi\Delta\varphi)dx=\int\limits_{\partial V}\Big(\psi\frac{\partial}{\partial n}\varphi-\varphi\frac{\partial}{\partial n}\psi\Big)dS$$
taking $\psi=\frac{1}{r}$ for which $\Delta\psi=-4\pi\delta(x)$.
A: Here's a probabilist's answer. Solving Laplace's equation with boundary conditions is equivalent to solving the following "gambler's ruin" problem.
A gambler has initial position ${\bf x} \in \mathbb{R}^3$, and moves inside the region as a(n isotropic) Brownian motion, until it hits the boundary of the box and ends the motion. There are two types of boundary faces: the "winning" faces, which has potential 1; and the "losing" faces, which has potential 0.
Question: Given the starting position ${\bf x}$, what is the probability $P({\bf x})$ that the gambler wins---i.e. hits the winning face(s) before the losing face(s)?
Observations:

*

*If ${\bf x}$ is on the winning face, then $P({\bf x})=1$.

*If ${\bf x}$ is on the losing face, then $P({\bf x})=0$.

*If ${\bf x}$ is anywhere else in the box, then a probability calculation shows that $P$ satisfies Laplace's equation: $\Delta P=0$.

With these in mind, we go back to the original question with $V_0=1$. The gambler starts at the origin ${\bf 0}$, and of the 6 boundary faces, 1 is winning and 5 is losing. By the symmetry (isotropy) of the problem, deduce that $P({\bf 0})=\frac{1}{6}$.
PS: If ${\bf x}\neq {\bf 0}$, the gambler's ruin story still holds, but we lose the symmetry argument that enabled us to get a quick answer.
