# Electric Field Energy and Potential of Sphere

1. Jun 17, 2012

### schaefera

1. The problem statement, all variables and given/known data
Using the definition of electric potential as well as the electric field energy definition of potential energy, determine the electric potential of a uniformly charged spherical shell of radius R.

2. Relevant equations
I know that V= -∫E.dl as we take dl from infinity to R. Also, I just learned how U=∫(1/2)*ε*E^2 dV, where we integrate over all space.

3. The attempt at a solution
From the first definition of potential, I can easily plug in E=Q/(4*∏*ε*R^2) and integrate to get V=Q/(4*∏*ε*R), which I obviously know to be the correct potential of the sphere.

Now, my problem with the second definition is that in finding U for the sphere, I get to the integral (1/2)*ε*Q^2/(16*π^2*ε^2) ∫ (1/r^2)*4*π*r^2 dr, where r is integrated from R to infinity (because E=0 inside the shell). When I evaluate this, I get Q^2/(8*π*ε*R), which is half of what I want... I don't get what's going wrong!

2. Jun 17, 2012

### TSny

Why do you think the answer should be twice what you got?

3. Jun 17, 2012

### BruceW

Um, just to check - are you allowed to assume that E=Q/(4*∏*ε*R^2) ? This is true for the electric field outside the shell, but I'm not certain if you are supposed to derive this or assume this. The question is a bit vague. Anyway, I'll assume that you are allowed to assume this.

Also, why have you used the symbol R in your equation for the electric field ? the shell has radius R, and your equation for the electric field is correct for values greater than R. I'm guessing you are just using the same symbol, even though you now mean that R can take any arbitrary value which is greater than the radius of the shell.

Ok, so your method was to integrate the electric field (outside the shell) to get the potential (outside the shell). It is almost correct, but not quite. There is something missing. Think about the limits of the integration.

4. Jun 17, 2012

### schaefera

Well, shouldn't these two methods yield the same answer aside from a factor of Q? That's why I think the second way should be twice what I get...

But after some googling, it seems like the potential energy of a spherical shell might have the 8 in the denominator, so why doesn't the potential found by the other method have an 8 (it has a 4, which came essentially from the fact that the electric field has a 4 in the denominator).

As for the limits, shouldn't I take it from R to infinity, because I need to do the integral over all space.

5. Jun 17, 2012

### schaefera

I think the weird thing comes from the fact that it's a conducting sphere, so it's almost like we're trying to find the potential energy of a spherical shell (all the charge is on the outside). But whereas I'd do an integral to find the potential energy of a nonconducting sphere (integrating over the entire sphere), it seems like you can't do an integral over just a shell...

6. Jun 17, 2012

### BruceW

Yes, sorry. You did it all correctly. I just confused myself over something.

Hmm. It's pretty interesting. The energy required to bring all the charges together to form the shell is the 'second definition'. And the 'first definition' is the potential energy (per unit charge) of a small test charge placed somewhere in the electric field created by the shell. So the two definitions are slightly different. Now, can you explain why the two definitions come up with these two different answers? (I am thinking about it myself, too).

EDIT: I think I know the answer now. Your problem was a pretty good problem, schaefera

Last edited: Jun 17, 2012
7. Jun 17, 2012

### BruceW

My advice is to think about what (mathematically) happens when you keep adding more charge to the shell, to build up the charge to the final amount.

8. Jun 17, 2012

### schaefera

I think I may understand now! Please let me know how this reasoning holds up.

So, let's start with the second definition: What was new here (to me, at least) was the idea that U=(integral)(1/2)(epislon)(E^2)... before I would always find potential energy by integrating with respect to dq (which I could then relate to r of the sphere). Using either of these methods, however, gets me the number I found in the second part of the problem. This is the potential energy of the sphere-- that is, the work that must be done to assemble the sphere of charge. Of course, you start out requiring zero work to add the first bit of charge, and then every time you add more charge you need to do more work due to the charge already present, which is essentially what the integral over dq stands for.

The first part, then, seems to refer to the potential (or, if I multiply by Q, the potential energy) of the very NEXT point charge that I bring near the already completed sphere. So, unlike the second part of the problem, this is an energy (or potential) calculated assuming the sphere already exists... the second part of the problem deals with actually creating the thing?

If that is right, though, it still bothers me that the method of part 1 can't be used to find the potential energy of the spherical shell itself. If (1/2)(epsilon)(E^2) is the field energy density (which I think I rightly interpret it to be) I'd imagine that integrating it over all space should give an agreement on the total U of the spherical shell...

Last edited: Jun 17, 2012
9. Jun 17, 2012

### TSny

Yes, the potential energy of a configuration of charges is equal to the work required to place the charges in that configuration (starting with the charges at some reference configuration which is usually taken to be the charges infinitely separated from one another).

You might be familiar with the formula for the energy of a charged capacitor: U = $\frac{1}{2}$QV. Note the factor of 1/2. You can write it as U = Q($\frac{V}{2}$) which can be interpreted as the charge times the average value of the potential while the capacitor was being charged from zero potential to its final potential V.

Anyway, I think your original answer for the energy is correct (with the 8 in the denominator).

10. Jun 17, 2012

### BruceW

Yeah, that's right. The potential is the energy required (per charge) to bring a charge from infinity onto the shell. So the small amount of work (dw) required to bring a small amount of charge (dq) onto the shell is related by dw=vdq And then from this, we can integrate to get the energy of the shell in terms of the charge on the shell.

From what you were saying, I guess you have used dw=vdq before, but because this question uses U instead, it was not obvious why that 1/2 appeared. It puzzled me too.

11. Jun 17, 2012

### BruceW

You can use the method for part one, but you had only calculated the potential. You then need to integrate with respect to charge to get the total energy. (Which agrees with the method of integrating the field energy density over all space).