The problem goes like this, from Purcell's electromagnetism:
Consider a charge distribution that has the constant density $ρ$ everywhere inside a cube of edge $b$ and is zero everywhere outside that
cube. Letting the electric potential $φ$ be zero at infinite distance
from the cube of charge, denote by $φ_0$ the potential at the center of
the cube and by $φ_1$ the potential at a corner of the cube. Determine
the ratio $φ_0/φ_1$.
In solving it, the author used dimensional analysis to say that the potential at the center has to be directly proportional to the whole charge divided by the side length of the cube and from there. I don't know why this is true. Can someone explain his insight and tell me how to generally use dimensional analysis to solve such problems?
Best Answer
The electric potential due to a point charge $q$ at the origin is (in SI units)
$$ \phi_\text{point}(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r} $$
This tells use that the product $\epsilon_0\phi$ must always have the same units as a charge per unit length. Which charge, and which length, are going to depend on the details of the integral you do over the charge distribution. The integral also might give you some stupid high-algebra factor like $\frac{2^5}{7}$ that depends on your geometry. But the dimensionful part of your integral is always going to come directly from the dimensionful part of your problem.
In this case you only have two parameters:
charge density $\rho$, with units $\text{charge}/\text{length}^3$
cube sizes $b$, a length
There is exactly one way to multiply these parameters to get charge per unit length, so every expression for a potential in this problem will be proportional to $\rho b^2$.
Beware that older E&M texts may use CGS units, rather than SI units. The beginner’s guide to CGS units for electromagnetism is that $4\pi\epsilon_0 \equiv 1$ is dimensionless. The advanced approach to CGS units is to replace $\epsilon_0$ with an expression involving the fine-structure constant. That’s kind of a rabbit hole; tread carefully.