Homework Help: Electric field energy sum due to surface charge?

1. Oct 23, 2006

redmage

http://img134.imageshack.us/img134/9600/emiv7.jpg [Broken]

The first integral is over the voume of a surface charged (q) sphere of radius a > R (radius of sphere) so a Gaussian surface beyond the sphere and the second is over the surface. I believe this sum would show the total electrostatic energy, since if a => infinity the surface integral goes to zero (as stated in my book) but I'm not 100% sure nor do I know how to show it is really more/less than the total if I'm wrong. If someone can tell me which it is and how to go about proving it I think I can take it from there, just don't know how to start. Thanks!

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2. Oct 23, 2006

quasar987

If you go back to the original formula from which you started, I believe it is something like

$$\frac{1}{2}\int \rho \Phi \ d\tau$$

You can see that where there are no charges, rho=0 and the integrand vanishes. So we can choose to integrate over only where there are charges, or over any volume whatesoever (as soon as it contains all the charges) and the integral will give the same result! You gotta make use of this fact to simplify the integral you wrote. If you integrate over all space, you should be able to show that the second integral vanishes!

3. Oct 23, 2006

redmage

So because I'm taking my integral beyond the surface of the sphere anyway, the whole thing reduces down to just the total energy no more, no less? Even though I'm adding the surface integral? That's where I'm still a little confused.

4. Oct 23, 2006

quasar987

What you wrote is the total energy. As soon as V encloses all the charges, what you wrote is the total energy. There's an infinity of volumes you can integrate over and it's still going to give you the correct value of the energy. But IN PARTICULAR, if you choose the volume to extend to infinity, the second integral vanishes, and you are left with a simpler expression of the energy.

5. Oct 23, 2006

redmage

Alright, I think I got it!

I have another related question, same assignment. I'm asked to find the surface charge density sigma of a two cylinder system (inner one with volume charge density rho) so the total energy is zero. I already derived the energy for the inner one, but I can't figure out what is the energy due to the surface charge density of the outer one. Also what does it mean to find the energy per unit lenght of the system? Wouldn't it be zero since the net energy is zero?

6. Oct 23, 2006

quasar987

Would you mind transcribing word for word what the question is asking please?

N.B. The total energy of a system is not the sum of the energy of its parts! Remember that the so called electrostatic energy is actually only the work necessary to assemble the system. I.e. the work done in bringing each charge one by one from infinity and glue it where it belongs in the system. So if you have a system made of two parts, and you assemble the first part first and then the second, then the work done in assembling the second part is influenced by field produced by the first part! So the work done in assembling the second system is not just the work you have to do against the field of the part you're currently assembling. You have to account for the field of the first part already assembled. Alright all that to say that total electrostatic energy is not "additive".

7. Oct 23, 2006

quasar987

A more mathematical way to see it is that if $\vec{E}_1$ is the field produced by part 1, and $\vec{E}_2$ is the field produced by part two, then the total field is $\vec{E}=\vec{E}_1+\vec{E}_2$. So the total electrostatic energy of the system is

$$\frac{1}{2}\epsilon_0\int_{\mbox{all space}}\vec{E}\cdot\vec{E}d\tau = \frac{1}{2}\epsilon_0\int_{\mbox{all space}}(E_1^2+2\vec{E}_1\cdot\vec{E}_2+E_2^2)d\tau$$

This is the energy of part 1 + the energy of part 2 + a cross term. This shows that the energy of a system is not the sum of the energy of its parts.

8. Oct 23, 2006

redmage

Oh I have that part ok, it's that I can't remember what/how to find E2 (the hollow cylinder with surface charge density)