How do one rigorously prove that the electric potential energy of an conducting sphere with charge $Q$ is $\frac{Q^2}{8\pi\epsilon_0R}$

Question

How do one rigorously prove that the electric potential energy of an conducting sphere with charge $Q$ is $\frac{Q^2}{8\pi\epsilon_0R}$? Is integration the only way?

Homogeneous charge distribution is assumed, I guess.

Answer 1

We will show that one need not calculate the energy stored in the electrostatic field by integration. To that end, we proceed.

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We start with the electrostatic energy $W$ stored as given by

$$W=\frac{\epsilon_0}{2}\int_V \vec E\cdot \vec E \,dV \tag 1$$

where $V$ is all of space.

We now assume that the field is induced by a charge distribution residing on a conductor of volume $V$ bounded by surface $S$. Inasmuch as the electric field is zero inside the conductor, the volume of integration in $(1)$ can be taken as all of space outside of the conductor.

We can change the form of $(1)$ using $\vec E=-\nabla \Phi$ along with the vector identity

$$\nabla \cdot \Phi\,\vec E = \Phi \nabla \cdot \vec E+\nabla \Phi \cdot \vec E\tag2$$

Then, we have for any conducting surface $S$

$$\begin{align} W&=\frac{\epsilon_0}{2}\int_V \vec E\cdot \vec E \,dV \\\\ &=-\frac{\epsilon_0}{2}\int_V \nabla \Phi \cdot \vec E \,dV \tag 3\\\\ &=\frac{\epsilon_0}{2}\int_V \Phi \nabla \cdot \vec E-\nabla \cdot (\Phi \vec E) \,dV \tag 4\\\\ &=\frac{\epsilon_0}{2}\oint_S\Phi (\hat n \cdot \vec E)dS \tag 5\\\\ &=\frac{\epsilon_0}{2}\left. \Phi(\vec r)\right|_{\vec r\in S}\frac{Q}{\epsilon_0} \tag 6\\\\ &=\frac12Q\left. \Phi(\vec r)\right|_{\vec r\in S} \tag 7 \end{align}$$

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NOTES:

In arriving at $(3)$, we used $\vec E = -\nabla \Phi$.

In going from $(3)$ to $(4)$, we used the vector identity $(2)$.

In going from $(4)$ to $(5)$, we used the Divergence Theorem, recognizing that the unit normal points out of $V$ and therefore into the surface of the conductor; we then absorbed the minus sign wherein the final result, the unit normal points out of the conductor's surface. We also used that Gauss's Law, $\nabla \cdot \vec E=\rho/\epsilon_0=0$ in $V$.

In going from $(5)$ to $(6)$, we used the fact that the potential is constant on the conducting surface and the integral form of Gauss's Law, $\oint \vec E\cdot \hat n dS=Q/\epsilon_0$.

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Thus, the electrostatic energy stored is given by

$$\bbox[5px,border:2px solid #C0A000]{W=\frac12 Q\left. \Phi(\vec r)\right|_{\vec r\in S}} \tag 8$$

For a conducting sphere, the potential on the surface is $\Phi = \frac{Q}{4\pi \epsilon_0R}$ which from $(8)$, immediately gives the result

$$W=\frac{Q^2}{8\pi\epsilon_0R}$$

as expected!

So, we see that one does not need to integrate $\vec E\cdot \vec E$ to determine the stored energy! Rather, we can simply use $(8)$ directly.

Answer 2

The work done in assembling the entire system, i.e, building it from radius $r\rightarrow r+ dr$ and the corresponding surface, to attain the full sphere, would be pretty rigorous, so that's an option.

Otherwise, you could always integrate $\frac{\epsilon}{2}\int_VE^2 dV$.

Answer 3

HINT:

Only two things needed:

1. The electric field generated by a charged sphere has rotational symmetry.

2. Gauss' law

From 1. and 2. you conclude that the field outside the sphere is the same with the field generated by all the charge concentrated at the center, and that the field inside the sphere is zero.