# Potential/self energy of a charged spherical shell

1. Feb 1, 2008

### Sourabh N

What is the potential energy of a uniformly charged spherical shell; i think it is called the self energy of shell.

2. Feb 1, 2008

### pam

The PE is QV/2.

3. Feb 1, 2008

### rohanprabhu

How is the potential energy independent of radius?? I'm getting something like:

$$U = \frac{\sigma^2 \pi r^5}{9\epsilon_0}$$

where, $\sigma$ is the charge density of the shell.

OR:

$$U = \frac{Q^2}{8\pi \epsilon_0 r}$$

Last edited: Feb 1, 2008
4. Feb 2, 2008

### pam

V=Q/4piepsilonzero r

5. Feb 2, 2008

### Ulysees

How low can the potential go = how much energy we can extract from the sphere.

Can we squash it?

Can we stretch it?

Last edited: Feb 2, 2008
6. Feb 3, 2008

### Shooting Star

??

The higher the potential energy, the more we can extract (if possible). If we squash it, we have to put in energy. By allowing it to stretch, we can get energy out of it.

7. Feb 3, 2008

### Ulysees

Whether you can get energy out or give energy to the system depends on whether the charges can move.

If it is insulating and rigid so charges cannot move, you can get or give 0 energy.

If it is conducting but rigid, charges can only move inside the conductor. Therefore you can only GIVE some energy to the charges by forcing them to redistribute under external force.

So what is the "self-energy" of the shell then? By convention it's the energy that the charges lose if they are allowed to repel each other to infinity. Only by convention. Potential energy can be set to whatever level you want, it's only the changes that matter, what you get or what you give.

8. Feb 3, 2008

### Ulysees

So the term "energy" is too often used loosely, we say things like "this object has this much energy" which is not strictly correct. Here's why.

Ill-stated question:

How much gravitational potential energy does the universe have?

9. Feb 3, 2008

### Lojzek

The second formula is right, the first one is wrong. The question is about a sphere, not a ball, so one must insert Q=4*Pi*sigma*r*r, so
the energy is:

E=2*Pi*r^3*sigma^2/epsilon

10. Feb 3, 2008

### Shooting Star

The total PE, calculated with the potential set to zero at infinity, can never be extracted, because we can never send all of the charge to infinity, but part of this total self-energy may be extracted. It's good that you have grasped the concept, because it was you who were asking the questions.
It can be broken up and the KE of the mutually repellent flying pieces may be collected to get some energy. Also, how did you get the charges in there in the first place? If that had required work to be done, perhaps the process can be reversed and the energy collected in a neater manner.

Just connect a wire to it and ground it, and collect part of the energy as the current flows.

Who's saying anything to the contrary?

I agree with you. But WHO exactly uses this meaningless question so often?

11. Feb 3, 2008

### Sourabh N

Well, thanks for the answers and comments, but was expecting the method to derive the equation.

12. Feb 3, 2008

### Ulysees

> Just connect a wire to it and ground it, and collect part of the energy as the current flows.

You know as well as I know that current means the charge has to go somewhere. It can't go to infinity in all directions. So you can't ever extract the full energy that we have (by convention) associated with the shell.

>> Ill-stated question: how much gravitational potential energy does the universe have?

> I agree with you. But WHO exactly uses this meaningless question so often?

You must have missed the point. It's the bit in bold ("have") Not whether the whole question profoundly concerns the whole of humanity or something. "How much potential energy" is like saying "how close together can masses come theoretically".

13. Feb 3, 2008

### Shooting Star

You never asked for the method...

The general formula for energy for a charge distribution is ½ [Integral over volume rho*V*(dxdydz)], where V is the potential at a point. To modify that for a spherical shell, we have to find W= ½ Integral over surface( s*V*da), where s is the surface charge density. The total charge is q and the radius is R., e = epsilon_nought.

V at the surface of the sphere = k*q/R, where k = 1/(4*pi*e). Then,

W = ½ kq/R Integral s*da = k/2(q^2/R) = [1/(8*pi*e)]*q^2/R.

14. Feb 3, 2008

### Shooting Star

In post #10, for that very reason, I have written "part" in bold italics.

Hey, I do get the point. I was just kidding a bit by asking WHO.

Cheers.