Irodov 3.138: Expanding Shell Spends Two Energies – Amit & Deepali Sinha Foundation

Problem Irodov 3.138

A thin spherical shell of radius $R_1$ carries a charge $q$ spread uniformly over it, and a point charge $q_0$ sits at its center. The shell expands until its radius is $R_2$ . How much work do the electric forces do during this expansion?

The shell of charge q at its initial radius R_1, the point charge q_0 at the center, and the dashed sphere of the final radius R_2, drawn for q, q_0 > 0. — The shell of charge $q$ at its initial radius $R_1$ , the point charge $q_0$ at the center, and the dashed sphere of the final radius $R_2$ , drawn for $q, q_0 > 0$ .

Given

shell charge $q$ , uniform
point charge $q_0$ at the center
radius $R_1 \to R_2$

Find

the work

A

done by the electric forces

Solution

Step 1Work as a drop in electrostatic energy

The electric forces are conservative, so the work they do equals the decrease of the system’s electrostatic energy $W$ :

A = W(R_1) - W(R_2) \tag{1}

Step 2The energy that depends on the shell radius

Two terms depend on $R$ . The shell’s own energy is half its charge times its own potential $q/4\pi\varepsilon_0 R$ ; the interaction energy is $q_0$ times the shell’s potential at the center, which is the same $q/4\pi\varepsilon_0 R$ everywhere inside.

\begin{aligned} W_{\mathrm{self}} &= \frac{1}{2}\,q\cdot\frac{q}{4\pi\varepsilon_0 R} = \frac{q^2}{8\pi\varepsilon_0 R}\\ W_{\mathrm{int}} &= q_0\cdot\frac{q}{4\pi\varepsilon_0 R} \end{aligned}

Adding them:

W(R) = \frac{q}{4\pi\varepsilon_0 R}\left(q_0 + \frac{q}{2}\right) + \text{const} \tag{2}

Step 3The work

Substituting (2) at $R_1$ and at $R_2$ into (1), the constant cancels:

A = \dfrac{q\,(q_0 + q/2)}{4\pi\varepsilon_0}\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)

(3)

Check

With $q_0 = 0$ the answer reduces to $(q^2/8\pi\varepsilon_0)(1/R_1 - 1/R_2)$ , which is exactly the field energy $\int \varepsilon_0 E^2/2\,dV$ of the shell’s own field $q/4\pi\varepsilon_0 r^2$ in the layer $R_1 < r < R_2$ , the region the field vacates. For like charges, $q q_0 > 0$ , and $R_2 > R_1$ , (3) is positive: repulsion does positive work as the shell grows, as it should.

Remark

The factor $1/2$ sits on the shell’s charge and not on $q_0$ . The shell’s charge feels the field of $q_0$ in full, but each element of the shell feels only the field of the other elements, which at the surface is the average of the shell’s field just inside (zero) and just outside. That half is the same factor that reappears as the pressure $\sigma^2/2\varepsilon_0$ on a charged surface. If $q_0 = -q/2$ the two effects cancel and the shell can expand or contract with no work at all.