# Electrostatic self-energy

1. Oct 12, 2013

### smantics

1. The problem statement, all variables and given/known data
Find an expression for the electrostatic self-energy of an arbitrary spherically symmetric charge density distribution ρ(r). You may not assume that ρ(r) represents any point charge, or that it is constant, or that it is piecewise constant, or that it does or does not cut off at any finite radius r. Your expression must cover all possibilities. Your expression may include an integral or integrals which cannot be evaluated without knowing the specific form of (r).

2. Relevant equations

$\phi$ = U/q

F=-∇U

E=-∇$\phi$

U = (1/2) $\int$ ρ$\phi$d3r

3. The attempt at a solution

For an arbitrary continuous charge distribution:

$\phi$(r) = $\frac{1}{4*pi*ε(0)}$ $\int$$\frac{ρ(r')}{abs(r - r')}$d3r'

For spherically-symmetric distribution of charge:

$\phi$ = ($\frac{1}{4*pi*ε(0)}$ 2$\pi$ ∫ sin$\theta$'d$\theta$' ∫ r'2dr' $\frac{ρ(r')}{r''}$ cos$\theta$)

Then solving for U:

U = (1/2) ∫ρ ($\frac{1}{4*pi*ε(0)}$ 2$\pi$ ∫ sin$\theta$'d$\theta$' ∫r'2dr' $\frac{ρ(r')}{r''}$ cos$\theta$) d3r

Could someone please tell me if I made any errors or assumptions that I should not have made for this situation?

Note: I couldn't figure out how to put limits on the integrals, but for the spherically-symmetric distribution of charge first integral (sin$\theta$'d$\theta$') is from 0 to pi and the second integral (r'2dr' $\frac{ρ(r')}{r''}$ cos$\theta$) is from 0 to infinity. These also affect the equation for U, and there is no limits on the first integral in U.

2. Oct 12, 2013

### TSny

Hello. I don't see where your factor of cos$\theta$ comes from in your last equation above.

If you're not required to use the equation U = (1/2) $\int$ ρ$\phi$d3r, I think there is an easier way.

Imagine bringing in successive thin spherical shells of charge from infinity and building up the charge distribution like layers of an onion. Thus suppose you have currently already built up a spherical distribution of charge of radius r. Find an expression for the change in potential energy dU in bringing in the next layer of charge of thickness dr. Then U = ∫dU.

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