Electrostatic Self-Energy of a uniform charged sphere

[AwesomeTrains]

Homework Statement

Hello, I have to calculate the self-energy of an uniform charged electron with radius R. The distributed charge is e.

Homework Equations

The SE is given as:
$E=\frac{1}{2}\int dV \int dV' \frac {\rho(\vec r)\rho(\vec r')}{ |\vec r - \vec r'|}$
according to the problem sheet.

The Attempt at a Solution

Since it's uniformly charged I guess $\rho(\vec r)=\rho(\vec r')=\frac{e}{(4/3)\pi R^3}$

And from here I'm stuck, I tried to evaluate the (ugly) integral: $\int_{0}^{2\pi} \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{R} \int_{0}^{R} \frac{e^2r'^2r^2sin(\theta')sin(\theta)}{2[(4/3)\pi R^3]^2\sqrt{(r-r')^2+(\theta-\theta')^2+(\phi-\phi')^2}} \, drdr'd\theta d\theta' d\phi d\phi'$ (In spherical coordinates.)

But maple just crashed when I put it in.
What am I doing wrong? Did I misunderstand the meaning of the prime? (Can I set $\vec r' = \vec 0$?)

Any hints are very appreciated.
Kind regards Alex

 

[ZetaOfThree]

It looks like you are approaching this problem in a very difficult, brute force way. Suppose you have a uniformly charge sphere of radius $r$. What is the energy required to add a thin, uniformly charge spherical shell of thickness $dr$ to it? See if you can use the answer to this question to help solve your problem.

 

[AwesomeTrains]

I think the work to add the shell is given by:
$W=\frac{1}{2} \int \rho(\vec r)\phi(\vec r)dr$
The potential $\phi(\vec r)$ is caused by the sphere with radius $r$. I think it can be viewed as a point charge potential, when seen from the outside. $(>r)$
Then I would add up these shells ($dr$) until I reach R.
Am I correct so far or am I confusing myself?

 

[ZetaOfThree]

$W=\frac{1}{2} \int \rho(\vec r)\phi(\vec r)dr$

I think that you have the right idea, but this formula is wrong. You should be able to see that the units are wrong. Recall that $dq=\rho dV$ where $dV$ is a differential volume element. So you need to find $dV$ for a thin spherical shell.

 

[AwesomeTrains]

That means r is constant:
$dV=r^2sin(\theta)d\theta d\phi$
If it's right I put it in W and integrate w from 0 to R over r?

 

[ZetaOfThree]

That's right. Remember to integrate over $\theta$ and $\phi$ as well.

 

[AwesomeTrains]

Thanks a lot for the help. Will evaluate the integral tomorrow and post the result, it's getting late :) NN

 

[AwesomeTrains]

Good morning, I'm stuck again :)
The work to add the spherical shell with radius $dr$ to the sphere with radius $r$:
$W=\frac{1}{2}\int\rho(\vec r)\phi(\vec r)r^2sin(\theta)d\theta d\phi$

Since it's uniformly charged: $\rho(\vec r)=\frac{e}{(4/3)\pi r^3}$
And the potential is: $\phi(\vec r)=\frac{e}{r}$
(I'm not sure about the charge, because we would end up with more than e charge on the sphere when we add all the shells to the sphere with radius $r$?)

The work is then: $W=\frac{1}{2}\int_{0}^{2\pi} \int_{0}^{\pi} \frac{e}{(4/3)\pi r^3}\frac{e}{r}r^2sin(\theta)d\theta d\phi=\frac{3e^2}{2r^2}$

Then I wanted to integrate through all the thin shells from $0$ to $R$ but $\int_{0}^{R} \frac{3e^2}{2r^2} dr$ is division by zero. Are the limits of the integral wrong?

 

[ZetaOfThree]

Since it's uniformly charged: $\rho(\vec r)=\frac{e}{(4/3)\pi r^3}$
And the potential is: $\phi(\vec r)=\frac{e}{r}$

Both of these are incorrect. If the sphere is uniformly charged, then $\rho$ is constant, i.e. independent of $r$ right? The potential is $\phi(\vec r)=\frac{q(r)}{r}$ where $q(r)$ is the charge of a uniformly charged sphere of radius $r$. $q(r) \neq e$!

 

[Mikhail_MR]

Both of these are incorrect. If the sphere is uniformly charged, then $\rho$ is constant, i.e. independent of $r$ right? The potential is $\phi(\vec r)=\frac{q(r)}{r}$ where $q(r)$ is the charge of a uniformly charged sphere of radius $r$. $q(r) \neq e$!

Correct: $\rho(\vec r)=\frac{e}{(4/3)\pi R^3}$ , R as radius. And $\phi(\vec r)=\frac{Q}{R} (\frac{3}{2}-\frac{r^2}{2R^2})$