Applying Gauss's Law Several Times

Doc Al

The field produced by a uniform shell of charge is zero at all points within the shell. A result derivable from calculus.

rude man

Just put Gaussian spherical surfaces around a < r < b, at b < r < c, and at r > c.
Inside a < r < b you have charge Q due to the inner sphere. Inside b < r < c you have zero total charge inside that surface since the E field all over that surface has to be zero. Look at my equation below. Inside r > c you have nothing but Q since the conductor is uncharged.

In all cases, total free charge inside a surface = 4πr^2 ε₀E.

Bashyboy

I'm sorry, I am still having difficulty with this. If I wanted to calculate the electric field at a point in between the insulated sphere and the conducting sphere--that is, I create a Gaussian sphere around the insulated sphere (a < r < b)--I would have the consider the electric field contribution of both the conducting sphere AND the insulated sphere, or would I just figure out the electric field at that point due to the enclosed charge in the Gaussian sphere, which would be the insulated sphere?

Doc Al

Because of symmetry, all you need is Gauss's law. That will tell you that the field is only due to the enclosed charge.

Bashyboy

By symmetry of what? the conductive sphere? How does that make the electric field at that point, due to the conductive sphere, zero?

Doc Al

Since everything in this problem is spherically symmetric, you can say that the field depends only on the distance from the center. Thus, for any spherical Gaussian surface, such as one drawn between a < r < b, you can treat the field as uniform. Thus you can apply Gauss's law to calculate the field. And that only depends on the charge within the surface, thus you can conclude that the contribution of charges external to the surface must be zero.

Bashyboy

Okay, I think I am beginning to understand. I believe I have just one more question, though. If we were to calculate the electric field for r > c, would we only have to consider the charge of the outer surface of the conductive sphere, because the net charge of the insulated sphere and inner surface of the conductive sphere is zero?

Doc Al

That is correct. (You of course consider all the charge within your Gaussian surface, which ends up being equal to just the charge on the outer surface of the conductor.)