# Charge inside conducting shell

1. Jun 24, 2011

### Shukie

1. The problem statement, all variables and given/known data
Inside a metal conducting shell of radius $R$ there is a negative charge $q$ at a distance $a$ from the center $M$. The shell is brought up to a potential $V$. What is the surface charge on the inside and outside of the conductor?

3. The attempt at a solution

I'm not sure if this is the way to solve this problem, but I know that in the case of a grounded conducting shell this system can be replaced by the original charge $q$ at distance $a$ from M and some imaginary charge $\frac{R}{a}q$ outside the shell, at distance $a' = \frac{R^2}{a}$ from $M$. Does the added potential change anything about this 'imaginary' setup?

2. Jun 25, 2011

### ehild

The method of mirror charges is used to calculate the field in case of a metal surface and a point charge. Here the surface charge is the question. Apply Gauss' Law.

ehild

3. Jun 25, 2011

### Shukie

I really don't know how to proceed. Do I add the potential of the negative charge and $V$ and calculate the electric field from that and then use Gauss' law to find $Q = \sigma A$?

4. Jun 25, 2011

### ehild

"The shell is brought up to a potential V." That should mean that the potential of the conducting shell is V (with respect to infinity, as no grounding is mentioned).

ehild

5. Jun 25, 2011

### Shukie

So we have:

$$V = - \int_{\infty}^R \mathbf{E} \cdot d\mathbf{r}$$

I don't know how to continue...

6. Jun 25, 2011

### ehild

What is the the potential of a spherical shell with charge Q at distance R of the centre?

ehild

7. Jun 25, 2011

### Shukie

Do I need to use image charges to solve this? Because otherwise I have no idea.

8. Jun 25, 2011

### ehild

Well, try the image charges. I do not see how you get the surface charge on the shell with that method, show me please. Note that not the surface charge density is asked, but the total charge both on the outer surface of a spherical conducting shell and on the inner one.

I would apply Gauss" law. You know it, do not you?

ehild

9. Jun 26, 2011

### Shukie

Yes, $\int \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\epsilon_0}$, but I don't know how to apply it here. The potential $V$ will it affect it right?

10. Jun 26, 2011

### ehild

Forget the charge inside the shell for a moment. There is a conducting spherical surface with charge Q. What is its electric field at distance r from the centre?

ehild

11. Jun 26, 2011

### Shukie

It would be 0 for $r < R$ and $\frac{Q}{4 \pi \epsilon_0 r^2}$ for $r \geq R$.

12. Jun 26, 2011

### ehild

OK. The electric field is the same as if all the charge were concentrated in the centre. What is the potential of such charged conducting sphere in terms of r?

ehild

13. Jun 26, 2011

### Shukie

That's $\frac{Q}{4 \pi \epsilon_0 r}$ for $r \geq R$.

14. Jun 26, 2011

### ehild

You know the potential at r=R. It is V. What is the charge then on the outer surface of the shell?
The position of the inner charge does not influence the field outside, as the metal wall screens it. The inner charge induces a charge distribution on the inner surface of the metal shell, and if you want to determine it, you can use the method of image charges. But you need only the total surface charge inside. You know that electric field is zero inside the metal wall. Taking a Gaussian surface inside the wall, the integral of E is zero on it, that is, the enclosed charge is zero. So what is the inner surface charge?

ehild

Last edited: Jun 26, 2011
15. Jun 26, 2011

### Shukie

So $Q = V 4 \pi \epsilon_0 R$ and so $\sigma_{out} = \frac{Q}{A} = \frac{V 4 \pi \epsilon_0 R}{4 \pi R^2} = \frac{V \epsilon_0}{R}$.

On the inner surface the charge is simply $q$, so $\sigma_{in} = \frac{q}{4 \pi R^2}$.

16. Jun 26, 2011

### ehild

No, σin, the inner surface charge density depends on the distance from the negative charge placed inside the shell. But the total surface charge is +|q|. If I understand the problem correctly, it asks only the surface charge and not its density.

ehild

17. Jun 26, 2011

### Shukie

Yes, you're right, so on the outer surfaceit's just $Q = V 4 \pi \epsilon_0 R$ and the inner surface $q$.

18. Jun 26, 2011

### ehild

Take care, the sign or the inner charge is opposite to that of q. It was said "q negative charge". The inner surface charge has opposite sign. So answer |q|.

ehild

19. Jun 26, 2011

### Shukie

Thanks a lot!

20. Jun 26, 2011

### ehild

You are welcome

ehild