What is the Electric Field Near a Small Hole in a Charged Spherical Shell?

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SUMMARY

The electric field near a small hole in a charged spherical shell can be analyzed using the principles of electrostatics. When a small patch is removed from a uniformly charged spherical shell with charge density sigma, the electric field just above and below the hole behaves similarly to that of an infinite plane. The field contributions from the remaining surface must be considered, but due to symmetry, the field above the hole is E/2, while the field below remains zero. This classic problem illustrates the significance of symmetry in electrostatic fields.

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wakko101
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The question : Consider a thin spherical shell of radius R with a uniform charge density sigma. If a very small piece of this surface were removed, leaving a small hole, what would the electric field be at a point just above/below the hole?

relevant info : the field due to the patch of surface (when intact) is the same as that of an infinite plane just below and above the surface (because if you're just above or below, the essentially flat surface looks like an infinite plane).

Does this mean that the field just above and below the surface is due solely to the patch itself, in which case the field would be 0? Or must one still take into account any contribution from all other points. Then again...if we're thinking about the points just above and below, would the leftover contributions be essentially perpendicular to the surface, in which case, it is 0 again? Did that even make sense?

This problem is confusing me, any hints/advice would be appreciated.

Cheers,
W. =)
 
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Try thinking of the hole as a point charge of the opposite charge...
 
This is an old classic problem.
When the surface is whole, the field on one side is E, the field on the other side is zero.
Using symmetry, tells you that E/2 is due to the small piece.
 

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