Electric Fields and Gauss' Law with a Conducting Sphere

[soccersquirt8]

Homework Statement

This isn't about a specific problem, but it is based off of a homework problem. There is an insulating sphere (from radius 0 to a), and it is concentric with a spherical conducting shell (from radius b to c). If I know the charge of the insulating sphere and the net charge OUTSIDE of the conducting shell, I should be able to find the charge at radius b and the charge at radius c.

Homework Equations

q_enc=epsilon*int(E dot dA)

The Attempt at a Solution

I know the spherical conducting shell must have E=0, which makes the charge at radius b equal to the negative of the charge in the insulating sphere. For clarity, I will say that the insulating charge has a charge q=-4, making the charge at radius b equal to +4. If the net charge equals -12 outside the conducting shell, then I have been told that the charge at radius c would be -8.

I can see that -8-4=-12, but I would think the charge in the insulating charge would play a part. I would think it would cancel out the charge at radius b as it did inside the conducting shell. If that were the case, then I would think the charge at radius c would be -12 because -12-4+4=-12, which is what I want. For it to be the other way like I was told, it seems like the -4 charge at radius b is acting twice, once to cancel out the +4 charge inside the insulating sphere and again to effect the charge at radius c. I drew electric field vectors outside of the conducting sphere, and I am only getting that the -4 charge canceling out the +4 charge, making the charge at radius c equal to the net charge. But apparently that is not right. Where is my line of thinking going wrong?

 

[SammyS]

The specific words used to describe the situation are VERY important.

Saying that "the net charge equals -12 outside the conducting shell" (I would take this to mean on the outer surface of the conducting shell.) says that you already know the charge on the outer surface (at r=c). ∴ The net charge on the conducting shell would be ‒12 + 4 = ‒8 .

However, if the textbook is saying that the net charge on the conducting shell is -12, the fact that the inner surface (r=b) has a charge of +q means that the charge on the outer surface is -16, because +4 + (-16) = -12 .

 

[soccersquirt8]

For all r>c, the net charge equals -12.

Going with your situation, where did the -4 charge factor in? It didn't seem like it played a part at all.

 

[SammyS]

For all r>c, the net charge equals -12.

Going with your situation, where did the -4 charge factor in? It didn't seem like it played a part at all.

It causes the charge on the inner surface of the shell (r=b) to be +4.