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#### jacobi

##### Active member

- May 22, 2013

- 58

- Thread starter jacobi
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- Thread starter
- #1

- May 22, 2013

- 58

- Thread starter
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- May 22, 2013

- 58

I'll post a solution, since I don't like unanswered threads.

The electrostatic self-potential energy is the amount of work required to increase the radius of a spherically symmetric charge distribution from r to r+dr. From the definition of voltage, \(\displaystyle dW=Vdq\). The charge dq is \(\displaystyle 4 \pi \rho r^2\). The potential V is \(\displaystyle \frac{kq}{r}=\frac{4 \pi \rho k r^3}{3r}=\frac{4 \pi \rho k r^2}{3}\). \(\displaystyle k\) is the Coulomb's Law constant, \(\displaystyle k=\frac{1}{4 \pi \epsilon_0}\). Multiplying these two expressions, we have \(\displaystyle \frac{16 \pi^2 \rho^2 k r^4}{3}\). Integrating from 0 to R (increasing the radius from 0 to R by putting on more and more shells), we have

\(\displaystyle \frac{16}{3} \pi^2 \rho^2 k \int_0^R r^4 dr=\frac{16}{15} \pi^2 \rho^2 k R^5\) as our answer.