Irodov 3.139: Pressure from a Shell's Energy – Amit & Deepali Sinha Foundation

Problem Irodov 3.139

A thin spherical shell carries charge spread uniformly over its surface with surface density $\sigma$ . Using conservation of energy, find the magnitude of the electric force that acts on each unit area of the shell.

The thin shell of radius R with its charge spread uniformly at surface density σ, drawn for σ > 0. — The thin shell of radius $R$ with its charge spread uniformly at surface density $\sigma$ , drawn for $\sigma > 0$ .

Given

uniform surface density $\sigma$ on a spherical shell

Find

the electric force per unit area

F_1

Solution

Step 1The energy of the charged shell

Call the shell’s radius $R$ and its charge $q$ . The shell is an equipotential at $\varphi = q/4\pi\varepsilon_0 R$ , so its electrostatic energy is half the charge times that potential:

W = \frac{1}{2}\,q\varphi = \frac{q^2}{8\pi\varepsilon_0 R} \tag{1}

Step 2A virtual expansion at fixed charge

The shell at radius R and, dashed, after a small expansion to R + dR. Each element dS is pushed outward by the electric force d F of the rest of the shell. — The shell at radius $R$ and, dashed, after a small expansion to $R + dR$ . Each element $dS$ is pushed outward by the electric force $d\vect F$ of the rest of the shell.

By symmetry the electric force on every element of the shell points radially outward and has the same magnitude $F_1$ per unit area. Let the shell expand slowly from $R$ to $R + dR$ with its charge $q$ unchanged. Energy conservation says the work done by these forces equals the decrease of the field energy (1):

\begin{aligned} F_1\cdot 4\pi R^2\,dR &= -dW = -\frac{d}{dR}\left(\frac{q^2}{8\pi\varepsilon_0 R}\right)dR = \frac{q^2}{8\pi\varepsilon_0 R^2}\,dR\\ F_1 &= \frac{q^2}{32\pi^2\varepsilon_0 R^4} \end{aligned}

Step 3In terms of the surface density

The charge is $q = 4\pi R^2\sigma$ ; substituting it, both $R$ and the numerical factors cancel:

F_1 = \frac{(4\pi R^2\sigma)^2}{32\pi^2\varepsilon_0 R^4} = \frac{16\pi^2 R^4\sigma^2}{32\pi^2\varepsilon_0 R^4}

F_1 = \dfrac{\sigma^2}{2\varepsilon_0}

(2)

The force is outward whatever the sign of $\sigma$ : like charges repel.

Check

Directly from the field: just outside the shell $E = \sigma/\varepsilon_0$ , just inside $E = 0$ , and a surface element feels the field of all the other charge, the average of the two, $\sigma/2\varepsilon_0$ . The force per unit area is then $\sigma\cdot\sigma/2\varepsilon_0$ , which is (2). Dimensions: $\sigma^2/\varepsilon_0$ is C $^2$ m $^{-4}$ over C $^2$ N $^{-1}$ m $^{-2}$ , that is N/m $^2$ , a pressure.

Remark

The result is the energy density $\varepsilon_0E^2/2$ of the field just outside the surface, with $E = \sigma/\varepsilon_0$ : the field pushes on a charged conductor with a pressure equal to its own energy density, a first look at the Maxwell stress. A soap bubble given a charge therefore swells, and for a fixed $\sigma$ the pressure does not depend on the radius at all.