# Work to remove point charge from center of spherical conducting shell

1. Oct 13, 2013

### paulharrylemon

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

A point charge q is at the center of a spherical conducting shell of inner radius a and outer radius b. How much work would it take to remove the charge out to infinity?

2. Relevant equations

Potential, W = 1/2qV

3. The attempt at a solution

I am going at this in layers, but getting the wrong result.
The work to bring the point charge to the center is zero (with nothing else in place yet). The work to bring in the -q on the inner radius a is W = -(1/2)qV.
The work to bring in the outer radius of charge q is W = (1/2)(q)[V from a + V from center point charge). The separation distance from a is (b-a) and just a from the center. Then what should I be doing with this? Should I add the work to bring in the -q to a and +q to b? When I do so, I get zero! :0 please help!

2. Oct 13, 2013

### TSny

Hello, paulharrylemon.

It can be a bit tricky to make sure you are taking care of all of the inter-particle interactions. For example, when you are bringing in -q to place at the radius a, there will be the work done due to interactions between -q and the central charge, and also there will be work done due to interactions between the elements of charge making up the total charge -q. That is, even if the central charge were not there, you would still have to do work in order to assemble the charge -q on the surface of a sphere of radius a.

Have you studied how to calculate electrostatic energy in terms of the electric field E distributed throughout space? If so, then that might be an easier way to get to the answer.

3. Oct 13, 2013

### paulharrylemon

Yes, and I did it that way. However our professor also wants us to figure it out this way :( thanks though :)

4. Oct 13, 2013

### TSny

OK. Well, looks like you will need to work out all the terms representing the various works. So, as you have said, you don't need to do any work to bring in the central charge.

Next: How much work is done in bringing in -q? Break this into two parts:
(1) The work that would have to be done if the central charge were not there
(2) The additional work done because of the presence of the central charge

5. Oct 13, 2013

### paulharrylemon

I am unsure of how to do part 1.
I am thinking of something similar to the work to build a solid sphere of charge, but don't know how to translate it to a surface. Any help? :)

6. Oct 13, 2013

### TSny

Note that once the charge is spread out over the sphere of radius a, then each charge element will be at the same potential V. Can you find the value of V at the surface of the sphere? Then you can use your formula W = U = (1/2)qV.

7. Oct 14, 2013

### paulharrylemon

Why are we looking at the surface of the sphere? We only have center charge and radius a at the moment, unless you just jumped ahead. Also confusing me is that shouldn't the surface of the sphere have the same V as at a since it's a conductor? Or is it just an equipotential with a different constant V? Thanks!

8. Oct 14, 2013

### TSny

I might be misunderstanding your approach to the problem. It appears to me from your previous remarks that you want to consider the work required to assemble the charge configuration which consists of a point charge q at the center of the conducting shell, a charge -q on the inner surface of radius $a$ of the conductor, and a charge q on the outer surface of radius $b$. The answer to the question would then be the negative of this work (since the problem essentially asks you for the work required to reverse this process.) I think this approach is ok.

So, forget the conductor for the moment and just consider the total work required to assemble the final charge distribution with all charges initially dispersed at infinity. Start by bringing in the point charge q. No work is required to do this. Now you want to bring in a total charge of -q and distribute it over a spherical surface of radius $a$. The charge -q is initially thought of as dispersed at infinity. You can bring in little elements of charge -dq from infinity and distribute these uniformly over the sphere of radius $a$ until you get the total charge -q spread uniformly on the surface. As you bring in each element -dq, there are forces between -dq and the central charge, q, as well as forces between -dq and the charge already placed on the sphere of radius $a$. So, you can think of the total work as the sum of the work done due to interactions of the elements -dq with the central charge and the work done due to interactions of the incoming elements -dq and the charge already on the sphere.

Finally, you can consider the total work required to bring in +q and distribute it over the sphere of radius $b$.

By the time you get all of the charges in place, the potential V on the surface of radius $a$ will be the same as the potential V on the surface of radius $b$. In fact, all points in between the two surfaces will be at the same potential. So, you can now fill up this space between the surfaces with a conducting material without any change in distribution of charge or any change in energy.

I don't think I understand this question.

9. Oct 14, 2013

### paulharrylemon

Thank you for the explanation :) I completely understand the logic behind it (thanks to you), but I just can't figure out how to translate it into math. I just handed it in, but I have understood every other homework problem this year so I'm not too worried! He gives solutions, so I'll post again if his explanation doesn't help!

10. Oct 14, 2013

### TSny

OK. Your method of assembling the charges starting with the charges at infinity will get the answer.

Another approach is to evaluate $U =\frac{1}{2} \sum q_i V_i$ over all charge elements of the system. In this approach, you are not considering assembling the charges, you are just evaluating the sum over the total charge distribution of the system when the point charge is at the center of the conducting shell.