# Electric potential puzzle

1. Aug 8, 2013

### BucketOfFish

This is question 2.30 from the 2nd edition of the Purcell book on Electricity and Magnetism. It's an interesting puzzle, and I've been thinking about it for a while, but I can't make any headway, so maybe you guys can do better.

Suppose you have a uniformly charged cube (charge density ρ) of side length b. The electric potential is set to 0 at infinity. What is the ratio of the potential at the center of the cube to the potential at a corner? Purcell claims you can solve this using only superposition, without requiring any mathematics. He also suggests you first consider a bigger cube which is 2b on each side.

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Here is what I have so far - due to symmetry, the potentials at the corners of a cube are identical. Due to superposition, this means that the potential at the center of any cube is equal to 8 times the potential at the corner of a cube which is half as long on each side. I can't prove that the corner/center potential ratio is independent of cube size, so I can't make any progress from here. Everything I tried ended up failing.

2. Aug 8, 2013

### Leveret

Say the argument you've given corresponds to a b^3 cube, subdivided into eight (b/2)^3 cubes. Would it still hold for a (2b)^3 cube split into b^3 cubes? Or a (cb)^3 cube split into (cb/2)^3 cubes, where c is any positive real number? Since your solution (i.e. 8) does not depend on b, you've already shown that it works for any cube.

3. Aug 8, 2013

### WannabeNewton

Hi Bucket! What you have to do is use some dimensional analysis.

First, let $b > 0$ be the side length of some cube. As you noted, because the potential is zero at infinity and the charge distribution throughout the cube is uniform, if we take any two points on the cube we can rotate the cube to swap the position of the points while leaving the cube invariant hence the entire system invariant meaning the potentials at the corners must all be equal; denote by $\varphi_1$ the corner potential and denote by $\varphi_0$ the central potential.

Now $\varphi_0 = f(Q,b)$ because $Q$ and $b$ are the only existing properties of an arbitrary cube hence they are the only things that can be used to characterize the potential at a fixed point in space due to an arbitrary cube. Now we know that $\varphi_0$ has to have the units of $\frac{q}{4\pi\epsilon_0 r}$; since the only length scale that $\varphi_0$ depends on is $b$, the only way it can have the above units is if $\varphi_0\propto \frac{Q}{b} \propto b^{2}$. Consider a cube of size $2b$; it is built up from 8 cubes of side $b$. Therefore, $\frac{\varphi_0(2b)}{\varphi_0(b)} = 4$ and by superposition $\varphi_0(2b) = 8\varphi_1(b)$ hence $\frac{\varphi_0(b)}{\varphi_1(b)} = 2$.

Last edited: Aug 8, 2013
4. Aug 8, 2013

### BucketOfFish

Thanks for the help guys, but I'm afraid that I'm still not fully convinced.

Leveret, you pointed out that a certain ratio holds regardless of cube size. However, this is the ratio of the center potential of a large cube to the corner potential of a cube eight times smaller. I feel it should be a small hop to use this fact to prove that the center/corner ratio of a single cube is also independent of size, but I can't figure out a way to do it.

Newton, I liked your dimensional analysis approach, but I don't really buy that the center potential of a cube is necessarily inversely proportional to b. What would rule out different relations like $\varphi_0 \propto \frac{Q}{\sqrt{b^2+3}}$, which would not result in the simple ratio you calculated?

By the way, numerical integration in Mathematica did show the result you calculated, so it does seem to be correct.

5. Aug 8, 2013

### WannabeNewton

What you wrote is not a valid expression because you can't add $b^{2}$ to $3$; the latter is unitless but the former has the units of $m^2$. That was my original point: the only length scale available is $b$ so the only way you can get the right units for the potential at a fixed point is if you have it proportional to $\frac{Q}{b}$. I will admit however that I myself am not 100% satisfied with this argument because I can't seem to make it fully mathematically rigorous; it is just something that seems intuitively correct from the standpoint of units. I will try to make it more rigorous but in the meantime if someone else has a way of making the above units argument more rigorous that would be great.

Anyways, there is a less elegant but more rigorous way to argue it. $\varphi_0 \propto \rho\int _{-b/2}^{b/2}\int _{-b/2}^{b/2}\int _{-b/2}^{b/2}\frac{dxdydz}{\sqrt{x^2 + y^2 + z^2}}$. Now define $x' = \frac{2x}{b}, y' = \frac{2y}{b},z' = \frac{2z}{b}$ then we have $\varphi_0 \propto \frac{\rho b^2}{4}\int _{-1}^{1}\int _{-1}^{1}\int _{-1}^{1}\frac{dx'dy'dz'}{\sqrt{x'^2 + y'^2 + z'^2}}$. The integral itself is just a constant value independent of the only existing properties $Q,b$ of the cube so we conclude that for fixed $\rho$, $\varphi_0 \propto b^2$.

6. Aug 8, 2013

### BucketOfFish

Wow thanks Newton! That was a really great answer!

7. Aug 8, 2013

### WannabeNewton

No problem! I might just end up making a HW thread on how to make the units argument in post #3 fully rigorous or at the least unequivocally convincing.