Irodov 3.140: Pulling a Charge Out of Its Cage – Amit & Deepali Sinha Foundation

Problem Irodov 3.140

An uncharged conducting spherical layer has inner radius $a$ and outer radius $b$ , and a small hole drilled through it. A point charge $q$ sits at its center $O$ . How much work must be done to carry $q$ slowly from $O$ out through the hole and away to infinity?

The uncharged conducting layer in section, inner radius a and outer radius b, with the charge q at its center O and the small hole on the right; the dashed arrow is the path out to infinity. Drawn for q > 0. — The uncharged conducting layer in section, inner radius $a$ and outer radius $b$ , with the charge $q$ at its center $O$ and the small hole on the right; the dashed arrow is the path out to infinity. Drawn for $q > 0$ .

Given

charge $q$ at $O$
radii $a$ (inner), $b$ (outer)
layer uncharged
small hole

Find

the work

A

to move

q

slowly to infinity

Solution

Step 1Work as a change in field energy

The charge is moved slowly, so it gains no kinetic energy, and the conductor’s charges stay in equilibrium at every stage. The external work then goes entirely into the electrostatic energy of the system:

A = W_{\mathrm{f}} - W_{\mathrm{i}} \tag{1}

Step 2The field before and after

The charges at the start: -q on the inner surface, +q spread evenly over the outer surface. The field fills the cavity and the outside, and the metal between a and b has none. The small hole is left out. — The charges at the start: $-q$ on the inner surface, $+q$ spread evenly over the outer surface. The field fills the cavity and the outside, and the metal between $a$ and $b$ has none. The small hole is left out.

At the start the field inside the metal is zero. Gauss’s theorem over a sphere inside the metal then puts $-q$ on the inner surface, and since the layer is uncharged, $+q$ on the outer surface, spread uniformly. In the cavity and outside, the field is that of $q$ alone at $O$ ; in the metal it vanishes:

E_{\mathrm{i}}(r) = \begin{cases} \dfrac{q}{4\pi\varepsilon_0 r^2}, & r < a \text{ or } r > b\\[2mm] 0, & a < r < b \end{cases}

At the end the charge is far away, the layer is uncharged and unpolarized again, and the field near $q$ is the bare field $q/4\pi\varepsilon_0 r^2$ at every distance $r$ from it.

Step 3The difference of the two energies

The energy density is $\varepsilon_0E^2/2$ . The two fields are the same everywhere except in the layer $a < r < b$ (measured from the charge), which held no field before and holds the bare field after. The infinite self energy of the point charge is the same in both and cancels, leaving

\begin{aligned} W_{\mathrm{f}} - W_{\mathrm{i}} &= \int_a^b \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_a^b \frac{dr}{r^2} \end{aligned}

Evaluating the integral and using (1):

A = \dfrac{q^2}{8\pi\varepsilon_0}\left(\dfrac{1}{a} - \dfrac{1}{b}\right)

(2)

Check

The work is positive, as it should be: once $q$ leaves $O$ , the charges it induces pull it back, and work must be done against them. A thin layer, $b \to a$ , gives $A \to 0$ , since a layer of no thickness takes no field away. The same number follows from potentials: at $O$ the induced charges give $\varphi_{\mathrm{ind}} = (q/4\pi\varepsilon_0)(1/b - 1/a)$ , and half of $q\varphi_{\mathrm{ind}}$ , with its sign changed, is (2), the factor $1/2$ because the induced charges are created by $q$ itself and fade as it leaves.

Remark

The answer is exactly the energy of the field of $q$ that the metal had excluded, the field in the spherical layer between $a$ and $b$ . A conductor lowers the energy of a nearby charge by pushing field out of its own volume, which is why a charge is always drawn toward an uncharged conductor; the thicker the closed metal around it, the more work it takes to get the charge out.