# Trouble with self-energy formula

1. Mar 7, 2014

### aguycalledwil

Hi everyone,

I'm having a little difficulty trying to calculate the self-energy of a uniformly charged sphere of radius a. That is, the work done in assembling such a distribution of charge. It is to my understanding that

0.5*integral(ρ*ø(r)dV)

should produce the answer, with ρ being the charge density (constant), ø(r) being some potential, and V being the volume of the sphere. However, I can't figure out what this potential is. Should it be the potential of a spherical shell, ie. Q/(4*pi*epsi0*[r^2])? I tried this but can't get the corrent answer out.

Will

2. Mar 7, 2014

### vanhees71

Of course, you have to solve for the electrostatic potential for the given charge distribution.

The potential obeys Poisson's equation (in Heaviside-Lorentz units)
$$\Delta \Phi=-\rho.$$
This we write in spherical coordinates. Due to symmetry we can assume that $\Phi=\Phi(r)$ is a function of $r$ only. That simplifies the equation to
$$\frac{1}{r} \partial_r^2 \left (r \Phi \right)=\rho.$$
For $r<a$ we have $\rho=\text{const}$. This gives
$$(r \Phi)''=\rho r \; \Rightarrow \; (r \Phi)'=-\frac{\rho}{2} r^2+C_1\; \Rightarrow \; \Phi(r)=-\frac{\rho}{6} r^2 + C_1+\frac{C_2}{r}.$$
Since the potential is continuous at the origin we must have $C_2=0$.

For $r>a$ we have $\rho=0$. There we have
$$\Phi(r)=C_1'+\frac{C_2'}{r}.$$
Since we want to define the potential to vanish for $r \rightarrow \infty$ we get $C_1'=0$. Further the total charge must be $Q=4 \pi a^3 \rho/3$ we find
$$C_2'=Q/(4 \pi).$$
Finally the potential must be continuous at $r=a$ which gives
$$-\frac{\rho}{6} a^2+C_1=\frac{Q}{4 \pi a}=\frac{a^2 \rho}{3} \; \Rightarrow \; C_1=\frac{a^2 \rho}{2}.$$
Integrating gives
$$E_{\text{em}}=2 \pi \rho \int_0^{a} \mathrm{d} r r^2 \Phi(r)=\frac{4}{15} \pi \rho^2 a^5.$$
You get the same result using directly the electric field, which is given by
$\vec{E}_r=-\Phi'(r) \vec{e}_r$:
$$E_{\text{em}}=\frac{1}{2} 4 \pi \int_0^{\infty} \mathrm{d} r r^2 \vec{E}^2.$$

3. Mar 9, 2014

### Philip Wood

By elementary methods, consider building up the sphere in thin shells. When the sphere's radius is r, the potential at its surface is $\frac{1}{4 \pi \epsilon_0 r} \frac{4}{3}\pi r^3 \rho$. So the potential energy of the outer shell (all at distance r from centre) is its charge multiplied by the potential we've just written, that is:
$$\frac{1}{4 \pi \epsilon_0 r} \frac{4}{3}\pi r^3 \rho \times 4 \pi r^2 dr\ \rho\ \ =\ \frac{4}{3} \frac{\pi}{\epsilon_0} \rho^2 r^4 dr.$$
So imagining the sphere to be assembled shell by shell, the total energy needed to do this is
$$\int_{0}^{a}\frac{4}{3} \frac{\pi}{\epsilon_0} \rho^2 r^4 dr \ = \ \frac{4}{15} \frac{\pi}{\epsilon_0} \rho^2 a^5.$$

4. Mar 9, 2014

### Staff: Mentor

No, you want the electrostatic potential (as a function of r) produced by a uniformly charged sphere, matching your original problem statement.

Finding this potential is a standard exercise in intermediate-level undergraduate E&M courses, and you will find a Google search on "potential of a uniformly charged sphere" to be fruitful. I see a whole group of hits on threads from Physics Forums alone!

5. Mar 10, 2014

### aguycalledwil

Thanks for the responses, these have been more than helpful!