Let $\mathbf{E}(\mathbf{x})$ be the electric field and $\varphi(\mathbf{x})$ the scalar potential due to a static charge density $\rho(\mathbf{x})$, with all quantities vanishing as $r=|\mathbf{x}|$ becomes large. The electrostatic energy of the configuration is given by

$$U=\frac{\varepsilon_{0}}{2} \int|\mathbf{E}|^{2} d V=\frac{1}{2} \int \rho \varphi d V$$

with the integrals taken over all space. Verify that these integral expressions agree.

Suppose that a total charge $Q$ is distributed uniformly in the region $a \leqslant r \leqslant b$ and that $\rho=0$ otherwise. Use the integral form of Gauss's Law to determine $\mathbf{E}(\mathbf{x})$ at all points in space and, without further calculation, sketch graphs to indicate how $|\mathbf{E}|$ and $\varphi$ depend on position.

Consider the limit $b \rightarrow a$ with $Q$ fixed. Comment on the continuity of $\mathbf{E}$ and $\varphi$. Verify directly from each of the integrals in $(*)$ that $U=Q \varphi(a) / 2$ in this limit.

Now consider a small change $\delta Q$ in the total charge $Q$. Show that the first-order change in the energy is $\delta U=\delta Q \varphi(a)$ and interpret this result.