I came across this problem in Electricity and Magnetism by E.M. Purcell:
A spherical shell of radius $a$ is charged with a uniform surface charge density $\sigma$. A small hole of radius $b << a$ is cut out (essentially a disk of radius $b$). What is the electric field at the center of the hole?
Intuitively, the direction of the field should be radially outward, though I'm having trouble finding it. I thought of plugging the disk back in to get a field of $\displaystyle\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{4\pi a^2\sigma}{a^2} \hat{\rho} = \frac{\sigma}{\epsilon_0}\hat{\rho}$, and then trying to "remove" the field due to the disk, but this doesn't seem to work. Any suggestions (the answer is $\frac{\sigma}{2\epsilon_0}\hat{\rho}$)?
Best Answer
I think you are on the right track. The idea is to consider the field of a complete, unpunctured sphere of surface charge density $\sigma$ and "add" a little patch of surface charge density $-\sigma$. By superposition, the patch and the full sphere is equivalent to a sphere with a small hole in it.
You are correct that the field right at the surface for the unpunctured sphere is $E=\sigma/\epsilon_0$. Now the trick I believe is that, for a point very close to the surface, the patch can be considered as an infinite plate of surface charge $-\sigma$. This is possible on the same grounds that one considers a plate infinite in extent if the point where we want to field is much closer to the plate than any of the physical dimensions of the plate. This seems justified here as you are at the centre of the "plate" and arbitrarily close to it. The field of such a plate is $-\sigma/2\epsilon_0$ so the net field at the centre is thus $$ \frac{\sigma}{\epsilon_0}-\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{2\epsilon_0} $$