Charge Distribution on Pear Shaped Conductor


My text book got a part about "point action", something like 'sharp objects will have more charges '. It asks if we are given a pear-shaped conductor (two sides with different radii of curvature) why charges accumulate more on the 'sharper side'? What thing is equal on the surfaces? How can we prove that 'sharp objects act as better charge emitters as in air-fresher'?


The thing that is equal all over the surface is the electric potential. Otherwise charges would be subject to forces that would rearrange them to make an equipotential surface. Equal potential requires that there be more charge on the regions of smaller radius. See Charge Distribution on Connected Spheres for an example.

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