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.

 

[rwisz]

I would approach this problem by first acknowledging that Gauss' Law is a fundamental principle in electrostatics and can be used to solve problems involving electric fields and charge distributions. In this case, we have a conducting sphere surrounded by an insulating sphere, and we are given the charge of the insulating sphere and the net charge outside the conducting shell.

To find the charge at radius b and c, we can use Gauss' Law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (qenc = ε0∫EdA). In this case, we can consider the conducting shell as our closed surface and use the fact that the electric field is zero inside a conductor to simplify our calculations.

We know that the electric field inside the conducting shell is zero, so the charge at radius b must be equal and opposite to the charge in the insulating sphere. This means that the charge at radius b is +4, as you correctly stated. However, we cannot simply subtract this charge from the net charge outside the conducting shell to find the charge at radius c. This is because the charge inside the insulating sphere also contributes to the electric flux through the conducting shell.

To solve for the charge at radius c, we need to consider the electric flux through the conducting shell due to both the charge outside and inside the shell. This can be calculated using Gauss' Law and setting the enclosed charge equal to the net charge outside the shell plus the charge inside the insulating sphere. This results in a charge of -12 at radius c, as you correctly calculated.

In summary, your line of thinking was correct, but you needed to consider the contribution of both the charge outside and inside the conducting shell when calculating the charge at radius c. It is important to remember that Gauss' Law takes into account all charges enclosed by a closed surface, not just the external charges.