# Do you think this makes sense?

1. Dec 17, 2009

### LucasGB

When a spherical gaussian surface is 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?

2. Dec 17, 2009

### kcdodd

I guess it just depends on what you define to be the surface of the sphere.

3. Dec 17, 2009

### LucasGB

If I define the surface of the sphere as the level I mentioned (leveled with the nuclei of the atoms of the outermost layer) then we can say there is a discontinuity in the electric field? Outside the sphere it falls with the inverse square, inside the sphere it is zero, and at the surface it is half the field at extreme proximity. Is this right?

4. Dec 20, 2009

### Staff: Mentor

No, since "at the surface" is not clearly defined.

Whenever you apply Gauss's law to a charged surface you'll get a discontinuity in the electric field. (Note that the charge is always contained within the Gaussian surface and not on it.) Consider an infinite plane sheet of charge. The field on one side is +σ/2ε0 and the other it is -σ/2ε0.

If that bothers you, then instead of arbitrarily modeling the surface charge on the sphere as being two surfaces, picture it as a uniform band of charge. That way you can imagine the field smoothly varying from zero to its full value outside of the sphere.