# Calculate the charge in a hollow spherical conductor

1. Feb 9, 2013

### Renaldo

1. The problem statement, all variables and given/known data

Consider a hollow spherical conductor with total charge +5e. The outer and inner radii are a and b, respectively.

(a) Calculate the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere. (Use the following as necessary: e.)

2. Relevant equations

Gauss's Law?

3. The attempt at a solution

I honestly don't have much of a clue. I feel like the problem should be pretty simple, but I'm not grasping all the aspects of Gauss's Law.

I hypothesized that the charge was 0 on the inner surface, and surmised that the charge on the outer surface should be equal to the total charge. This was wrong. I then hypothesized that the inner charge would be -3e and that the outer charge would need to be added to this to result in 5e, so 8e. Also not correct.

2. Feb 9, 2013

### haruspex

Close, but you made a small mistake. Think that through again.

3. Feb 9, 2013

### Renaldo

The inner surface would be +3e because the positive charge would be attracted by the negative field, accumulating on the outside of the sphere. The charge on the outer surface would be +2e because the field is weaker there, and we know the total charge is +5e. Is this correct?

4. Feb 9, 2013

### haruspex

Sounds right to me. But it also sounds as though you are basing this on a hunch rather than on known principles. Consider the field in the conductor, i.e. between the inner and outer surfaces. What do you know about that? How would you calculate it from the isolated charge in the cavity, the charge on the inner surface, and the charge on the outer surface?

5. Feb 9, 2013

### Renaldo

All right, I will give it a shot.

q/εo = ∫Eda
q/εo = E4∏r2

E = kq/r2

E1 = kq1/b2

E2 = kq2/a2

E1+E2 = kq1/b2 + kq2/a2

E3 = E1+E2
E3 = kq3/r2

kq3/r2 = kq1/b2 + kq2/a2

This leaves me with three variables.

6. Feb 9, 2013

### Dick

The electric field inside of a conductor is zero, right? Pick a radius r where a<r<b. So it's inside of the conductor. Then use Gauss' law at radius r. What does that tell you?

Last edited: Feb 9, 2013
7. Feb 9, 2013

### Renaldo

Perhaps I am confused as to the nature of hollow spherical conductors. If I understand the problem correctly, a point where a<r<b would be inside the metal of the shell. What is the purpose of using Gauss's Law here? Would not the field at this point be = to 0?

8. Feb 9, 2013

### Dick

Yes, it would. So the enclosed charge must be zero, yes?

9. Feb 9, 2013

### Renaldo

(a) Calculate the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere.

So the enclosed charge is not equal to -3e? What does -3e represent?

10. Feb 10, 2013

### Dick

The enclosed charge is the sum of -3e plus the charge on the inner surface of the conductor. The Gauss integral is zero. Hence, what's the charge on the inner surface on the conductor? Help me out here.

11. Feb 10, 2013

### Renaldo

Oh, I see. The enclosed charge is equal to the charge on the inner radius, b, plus the charge in the center. It's obviously 3e, since the field is equal to 0.

As for the outer radius, I know that the spherical conductor has total charge, +5e. In order for it to have this total charge, I add the charges on the two surfaces together:

3e + 2e = 5e

It's really quite simple, but I was failing to understand the logic behind it. Thanks for the help.