E-field on a spherical shell with a hole

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A uniformly charged spherical shell with surface charge density \sigma contains a circular hole in its surface. the radius of the hole is small compared with the radius of the sphere. what is the electric field at the center of the hole?

I only know this question requires superposition of the sphere and the circle but how do we do it? I thought we can only find the e-field at some point due to a charge? Are we suppose to consider infinitesimal charge on the sphere and do a integration over the entire surface to find the e-field on the cavity? Also, does the size of the hole affect the answer?
 
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The key here is that the radius of the hole is small compared to the radius of the sphere. Using that approximation, it is reasonable to assume that the hole is more or less flat.

Do you know how to calculate the field at the center of a uniformly charged disk, carrying surface charge density -\sigma? What do you get when you superimpose such a disk on a much larger spherical shell carrying surface charge density +\sigma?
 
How do you do that? I only know how to calculate the field at a distance surrounding a charge. I suppose we have to use Gauss's law?
 
If you don't know how to calculate the field of a flat circular disk, at a point on its axis or at its center, you need to open your textbook. That calculation is done as an example in almost every introductory EM text I can recall seeing.
 
Actually I meant how do you calculate the field of the solid sphere. Do you mean calculating the field on the surface of the sphere as if there is a charge at the center of the sphere?
 
There are several methods for finding the field of a uniformly charged spherical shell. The easiest method is to use Gauss' Law...can you think f what symmetries a field produced by a uniformly charged spherical shell would have, and a Gaussian surface you can use to exploit those symmetries?

You will find that the field is discontinuous at the surface, and so you might consider using the average of the fields just inside and just outside the shell for this problem, since when the hole is present, there will be no discontinuity there.
 
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