- #1

LucasGB

- 181

- 0

*very*close to the surface of a charged sphere, but not

*at*the surface of the sphere, the electric field is, according to Gauss' Law:

Q/(epsilon 0 times area of the surface)

where Q is the charge of the sphere. But when the gaussian surface is

*at*the surface of the sphere, half the charge is inside the surface, and half the charge is outside. (Think of a line of atoms with orbiting electrons. The gaussian surface is leveled with the nuclei, so statistically speaking, half the electrons are above it, and half are below it.) Therefore, according to Gauss' Law, the electric field is:

Q/2(epsilon 0 times area of the surface)

Which is half the electric field at extreme proximity. Inside the sphere the field would, of course, be zero. Does this seem sensible to you?