Irodov 3.137: The Field Pays for Expansion – Amit & Deepali Sinha Foundation

Problem Irodov 3.137

A spherical shell of radius $R_1$ carries a charge $q$ spread uniformly over it. The shell is expanded to radius $R_2$ . Find the work done by the electric forces during the expansion.

The shell of radius R_1 about O with its uniform charge q (drawn positive), and dashed, the radius R_2 to which it is expanded. — The shell of radius $R_1$ about $O$ with its uniform charge $q$ (drawn positive), and dashed, the radius $R_2$ to which it is expanded.

Given

uniformly charged spherical shell, charge $q$
initial radius $R_1$
final radius $R_2$

Find

the work

A

of the electric forces

Solution

Step 1The energy of a charged shell

By symmetry the field of the shell is radial and the same all round; it vanishes inside, and outside Gauss’s theorem $E\cdot4\pi r^2 = q/\varepsilon_0$ makes it the field of a point charge. Integrating the density $\varepsilon_0E^2/2$ over spherical layers outside a shell of radius $R$ :

\begin{aligned} W(R) &= \int_R^\infty\frac{\varepsilon_0}{2}\left(\frac{q}{4\pi\varepsilon_0 r^2}\right)^2 4\pi r^2\,dr = \frac{q^2}{8\pi\varepsilon_0}\int_R^\infty\frac{dr}{r^2} \end{aligned}

W(R) = \frac{q^2}{8\pi\varepsilon_0 R}

(1)

Step 2The work as the energy given up

The electric forces are conservative, so their work equals the decrease of the electrostatic energy between the initial and the final state. The charge stays uniform throughout, so (1) applies at both radii:

A = W(R_1) - W(R_2)

A = \frac{q^2}{8\pi\varepsilon_0}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

(2)

For $R_2 > R_1$ the work is positive: the shell’s charges repel one another, and the electric forces push the expansion along.

Check

Direct route through the force on the shell. The field just outside is $q/(4\pi\varepsilon_0 R^2)$ and inside zero, so the surface charge feels the average, and the outward pressure is $p = \sigma E_{\text{out}}/2$ with $\sigma = q/(4\pi R^2)$ ; the work in a step $dR$ is $p\cdot 4\pi R^2\,dR$ :

A = \int_{R_1}^{R_2}\frac{q}{4\pi R^2}\cdot\frac{q}{8\pi\varepsilon_0R^2}\cdot4\pi R^2\,dR = \frac{q^2}{8\pi\varepsilon_0}\int_{R_1}^{R_2}\frac{dR}{R^2},

which is (2).

Remark

The energy released is exactly the field energy that filled the space between $R_1$ and $R_2$ : expansion wipes out the field there and leaves the field outside $R_2$ untouched. Expanded to infinity, the shell hands over all of its self energy $q^2/(8\pi\varepsilon_0R_1)$ as work, which is one way to see that the field energy is real: it can be cashed in as mechanical work.