A spherical drop of water has radius R = 5 * 10^-4 cm. The drop has an even surface charge distribution.
Consider an element dA of the surface. This element is exposed to a force from the rest of the surface elements.
The effective field in the surface is equal to the mean value of the effective field just outside and just inside the surface.
Use Gauss's Law to compute the effective field in the surface.
The Attempt at a Solution
I've done several Gauss's Law exercises before, but in this one I'm not even sure I understand the question.
What bothers me is the line about the "effective field" being the "mean value" of the field just inside and just outside the surface. In one problem I just did, I showed that the electric field inside an even spherical surface distribution is zero, so how can the field on the surface be the "mean" of 0 and a field slightly less than on the surface?
Am I misreading something?
Appreciate any help on this one!