# Determining total charge on the surfaces of spherical conductors

In the attached picture is all of the information to complete this problem. The picture is of a solid sphere at the center of a hallow sphere, both of which are conductors.

The question asks to find the total charge of the exterior and interior surfaces of the hollow conductor, as well as the exterior surface of the solid conductor...

I've attempted using Gauss' Law in various ways, but nothing seems to work, and I'm unsure of how to calculate the net E-field to 15,000 N/C, as indicated in the picture. If someone could point me in the right direction it's be greatly appreciated! :]

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Gah. Just as I was about to head to bed. I can't get into much detail until the morning, but the advice I can give is to remember your equation for the field due to a spherical charge distribution as well as 2 important rules:

1. In a spherical distribution the excess charges on a conductor align uniformly on the outer surface.
2. The charges will be aligned such that there is no net electric field inside the conductor itself (although there may be a net field in the gap between the inner and outer conductors if there is a net charge on the inner conductor).

Choosing Gaussian surfaces to take advantage of these properties, you should be able to determine the signs (do this first) and magnitudes of the net charges at play. Hope this helps. If you're still unclear on where to go, I'll return to this thread in the morning.

Hmm....then it would seem to me that the point outside both the spheres would be easiest to calculate! By what you've said...do you mean that the inner surface of the Hallow sphere is actually inducing no E-field at that point..?

I think I am only confused now about the Inner E-field, I will attempt the Gaussian surfaces at these points and see how they go...thanks!

Wow...I got it!! The inner charge has to be opposite the inner charge, and I was able to solve for the two points by utilizing gaussian surfaces to give me the charges of the exterior surface of both! Your help is greatly appreciated!

WOO!