# Potential of a conducting sphere in a conducting shell

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1. Apr 9, 2017

### razidan

1. The problem statement, all variables and given/known data
A conducting sphere, radius R, charged with Q is inside a conducting shell (2R<r<3R) with charge 2Q. Find the electric potential and the energy.

2. Relevant equations
$\Phi =-\int_{r_1}^{r_2} \vec{E}\cdot\vec{dl}$
$U=\int_{V}E^2dV$

3. The attempt at a solution
I think i got it right, and i'm mostly looking for confirmation:
I started with calculating the field everywhere:
$\vec{E} (r) = \begin{cases} 0 & \quad \text{if } \text{ r<R}\\ \frac{kQ}{r^2} \hat{r} & \quad \text{if } \text{ R<r<2R}\\ 0 & \quad \text{if } \text{ 2R<r<3R}\\ \frac{3kQ}{r^2} \hat{r} & \quad \text{if } \text{ r>3R}\\ \end{cases}$

$\Phi (r) = \begin{cases} \frac{kQ}{2r} & \quad \text{if } \text{ r<R}\\ \frac{3kQ}{3R}+\frac{kQ}{2R}-\frac{kQ}{r} \hat{r} & \quad \text{if } \text{ R<r<2R}\\ \frac{3kQ}{3R} & \quad \text{if } \text{ 2R<r<3R}\\ \frac{3kQ}{r} \hat{r} & \quad \text{if } \text{ r>3R}\\ \end{cases}$
$\text{and the energy is} \quad 14\pi\frac{k^2Q^2}{R}$

Thanks,
R

2. Apr 9, 2017

### John Morrell

There are a couple things that I'm unclear about. What do you mean when you say 'energy'? Do you mean the potential energy of a particle far away from the sphere? This seems a bit unclear. Also, I'm assuming for the first part that you are measuring the potential difference from infinitely far away; there is no real meaning to an absolute electric potential, it all depends on where you take it in reference to. Is there any more information that came with the problem?

I don't know if this was a typo, but when you list the potential energy for r<R you list it as varying with radius r. That shouldn't be the case; the potential within a conducting body should always be constant. A conducting volume is a large equipotential region; the way you have it listed would give an electric field within the sphere of kQln(2r), which is obviously not true.

3. Apr 9, 2017

### razidan

Hello and thanks for the response:
what I mean by energy is the electrostatics energy needed to construct such a system, by bringing infinitesimal amounts of charge, bit by bit from infinity.
and about "r" - it is indeed a typo... for some reason i cannot edit that post.

R

4. Apr 9, 2017

### haruspex

I agree with your fields but not your potentials. How did you arrive at them?

5. Apr 9, 2017

### razidan

Thanks, missed a "-" sign:
$\Phi (r) = \begin{cases} \frac{3kQ}{2R} & \quad \text{if } \text{ r<R}\\ \frac{3kQ}{3R}-\frac{kQ}{2R}+\frac{kQ}{r} & \quad \text{if } \text{ R<r<2R}\\ \frac{3kQ}{3R} & \quad \text{if } \text{ 2R<r<3R}\\ \frac{3kQ}{r} & \quad \text{if } \text{ r>3R}\\ \end{cases}$

6. Apr 15, 2017

### razidan

A follow up question (given that the potential in my last response is correct):
The sphere and the thick shell are now connected with a conductive wire. what is the potential and energy now?
This part has me stumped. I know that now the sphere and the shell have the same potential because charged can transfer.
But i'm not sure what to do.

Thanks,
R

7. Apr 15, 2017

### haruspex

It was. Sorry, I omitted to confirm.
Right. What is conserved? What equations can you write relating the charges and the potentials?

8. Apr 15, 2017

### razidan

I understand that the charge is being conserved.
$Q_1 +Q_2 =3Q$
I'm not sure that writing this is correct, though:
$\Phi(r<R)=\Phi(2R<r<3R) => \frac{3kQ_1}{2R} = \frac{3kQ_2}{3R}$

I'm not sure if this is correct because the numerical factors all came from the integration of the fields, before the charge was transferred...

9. Apr 15, 2017

### haruspex

Quite so, you cannot use those. So recalculate the fields and potentials in terms of two unknown charges.

10. Apr 15, 2017

### razidan

Ok, I will try this, Thanks!

11. Apr 15, 2017

### razidan

I have found that the charge on the inner sphere is now zero. Is that correct? If so, Is is because this configuration has a lower energy associated with it, and so the system "prefers" this configuration?

12. Apr 15, 2017

### haruspex

Yes.

13. Apr 15, 2017

### razidan

Thank you very much!