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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 $\frac{\sigma}{2\epsilon_0}\hat{\rho}$)?

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    $\begingroup$ It's been a while since I did electrostatics, so apologies if this isn't right. But here's my idea (which is very similar to your suggestion and seems to give the right answer) Consider the following complementary problems: a spherical shell of radius $a$ and charge density $\sigma$ (no hole), and a small circular disc of radius $b$ and charge density $-\sigma$, then evaluate the electric fields individually. What happens if the disk is placed on the shell? $\endgroup$
    – Philip
    Commented Jul 24, 2017 at 23:14

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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 $$ \frac{\sigma}{\epsilon_0}-\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{2\epsilon_0} $$

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  • $\begingroup$ Thank you - to reiterate, since the distance from the plate is "infinitely small", one can use Gauss's Law to compute the electric field due to the disk as: $\oint E \cdot \, d\vec{A} = \frac{\sum Q}{\epsilon_0} \Rightarrow (E)(2\pi r^2) = \frac{\pi r^2 (-\sigma)}{\epsilon_0} \Rightarrow E = \frac{-\sigma}{2\epsilon_0}$, so the total is just $\frac{\sigma}{\epsilon_0} - \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{2\epsilon_0}$? $\endgroup$
    – Shreyas B.
    Commented Jul 24, 2017 at 23:47
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    $\begingroup$ The first key point is, to compute the effect of the puncture, one consider the small hole to be very small. The second key point is that, to compute the field at the centre of this very small disk, the small disk might was well be infinite in extent since what matters is the ratio of the distance to this disk/radius of the disk, a quantity that will go to $0$ for any $b\ne 0$. As for the rest, yes it's just Gauss's law as per your comment. $\endgroup$
    – ZeroTheHero
    Commented Jul 24, 2017 at 23:51
  • $\begingroup$ What I don't understand is this: the potential at the center of a disk of radius $a$ can be computed by considering concentric rings of width $dr$ and radius $r$. Since $\displaystyle d\phi = \frac{1}{4\pi\epsilon_0}\frac{\sigma\, dx\, dr}{r}$, where $x$ represents $\phi$ for all the pieces on the ring of radius $r$, we can integrate twice to get that $\phi = 0$ at the center of the disk. If $\displaystyle\vec{E} = -\nabla\phi$, that implies $\vec{E} = 0$. What is wrong here? $\endgroup$
    – Shreyas B.
    Commented Jul 25, 2017 at 0:11
  • $\begingroup$ I'm not sure using the potential is productive since the geometry makes it hard to compute $\phi$ anywhere except on the symmetry axis of the plate. This is not enough to compute the gradient since in principle the potential could have non-zero $\partial/\partial r$. $\endgroup$
    – ZeroTheHero
    Commented Jul 25, 2017 at 1:37
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    $\begingroup$ Actually, the reason I didn't write an answer to this is because there is something to what @ShreyasB. is saying that I couldn't explain clearly. I believe the field at the center of a charged disk is zero. Take, for example, this answer, which shows that $E = \sigma/2\epsilon_0$ works only for $z>0$. Directly at the center $z=0$, all the contributions cancel out and give you $E=0$. The solution, I think, is that an infinitely thin sheet is itself quite unrealistic and so you could just consider yourself to be infinitesimally close to it. $\endgroup$
    – Philip
    Commented Jul 25, 2017 at 22:07

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