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Electrostatic self-potential energy

jacobi

Active member
May 22, 2013
58
Find the electrostatic self potential energy of a spherical charge distribution with charge density \(\displaystyle \rho\) and radius \(\displaystyle R\). The self potential energy is the work required to increase the radius of the sphere from \(\displaystyle r\) to \(\displaystyle r+dr\).
 

jacobi

Active member
May 22, 2013
58
Wow, I forgot this was even here.
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.