# Work moving point charge from center of conducting shell

1. Sep 15, 2015

### Jonathan K

1. The problem statement, all variables and given/known data
A point charge q is at the center of an uncharged spherical conducting shell, of inner radius a and outer radius b. How much work would it take to move the point charge out to infinity (through a tiny hole drilled in the shell)?
[answer: q2/8πε0)(1/a - 1/b)

2. Relevant equations
W = ε0/2 ∫E2

W = 1/2 ∫ρV dτ

3. The attempt at a solution
Since the point charge induces a uniform charge of -q on the inner surface and q on the outer surface, I figured I could figure out the total energy of this system and then this would be how much energy you get back when dismantling it by taking q out infinitely far away. So I figured it takes zero work to initially move the point charge and then I calculated the energies of the two induced surface charges using the second integral above but I get
W = q2/8πε0 (1/a +1/b). What' wrong here?

2. Sep 15, 2015

### BvU

Hi Jon, welcome to PF !

What do you get for the energies from the two surface charges ?

When I calculate the energy in the field in all of space after and before and then subtract, I only have to integrate your first formula from a to b . The minus sign from the subtraction swaps a and b.

I see you didn't get a reply for your first post. Sorry about that. No useful expertise, but I'll look into it.

3. Sep 15, 2015

### Jonathan K

For the inner surface I got q2/8πε0a and the outer q2/8πε0b. As for your calculation, wouldn't the total energy after be just zero since the charge is infinitely far away from a neutral shell? I wasn't entirely sure how to use the first integral since we have two different electric fields: one inside the conducting material and one outside.

4. Sep 15, 2015

### TSny

Did you take into account that the induced charge on the inner surface has opposite sign from the charge on the outer surface?

When the point charge is removed to infinity, there is still the energy of the field of the point charge. This energy is infinite! (This is the infamous infinite "self energy" of a point charge.) However, that infinite energy is also present in the initial configuration. Fortunately, you only need the difference in initial and final energies. So, the infinite self energy cancels out and you just get the integral of the squared field between a and b, as noted by BvU.

5. Sep 15, 2015

### Jonathan K

I did take into account the sign difference; I found that the potential on the inner surface and charge density were both negative so the negative cancelled when evaluating the integral. I'm a bit confused by why you would integrate the field squared between a and b since that is space filled by conductor where there is no field.

6. Sep 15, 2015

### BvU

Precisely ! So before there is this space with no energy from the E field. The rest of space has a field that is the same as if there were no conductor.
After you have the entire space filled with field. The difference is the field energy that has to be supplied. I.e. the work needed to move the charge to infinity.

Last edited: Sep 15, 2015
7. Sep 15, 2015

### Jonathan K

Ahh I get it now, the field energy in the space between the inner and outer surfaces is the only difference in energy from the before and after situations. Thanks for your help!

8. Sep 15, 2015

### TSny

Using the formula $W = \frac{1}{2} \int \rho(\vec{r}) V(\vec{r}) d^3r$ is a little tricky.

First, you can think of the charge density $\rho(\vec{r})$ as the charge density, $\rho_{p}$, due to the point charge $q$ at the center, plus the charge density on the inner surface, $\rho_{in}$, plus the charge density on the outer surface, $\rho_{out}$.

So $W = \frac{1}{2} \int \rho_p(\vec{r}) V(\vec{r}) d^3r + \frac{1}{2} \int \rho_{in}(\vec{r}) V(\vec{r}) d^3r + \frac{1}{2} \int \rho_{out}(\vec{r}) V(\vec{r}) d^3r$.

Show that the last two integrals cancel each other.

You are left with $W = \frac{1}{2} \int \rho_p(\vec{r}) V(\vec{r}) d^3r$. This can be broken up into three separate integrals by writing $V(\vec{r})$ as the superposition of the potential at $\vec{r}$ due to the point charge alone, $V_p(\vec{r})$; the potential at $\vec{r}$ due to the charge on the inner surface alone, $V_{in}(\vec{r})$; and the potential at $\vec{r}$ due to the charge on the outer surface alone, $V_{out}(\vec{r})$. Thus,

$W = \frac{1}{2} \int \rho_p(\vec{r}) V_p(\vec{r}) d^3r + \frac{1}{2} \int \rho_p(\vec{r}) V_{in}(\vec{r}) d^3r + \frac{1}{2} \int \rho_p(\vec{r}) V_{out}(\vec{r}) d^3r$

The first integral is the pesky infinite self energy. But this energy will still be there after the point charge is removed from inside. So, it cancels when considering the change in energy.

So, you just have to evaluate the last two integrals.

Last edited: Sep 15, 2015