Irodov 3.64: Only the Totals Reach the Center – Amit & Deepali Sinha Foundation

Problem Irodov 3.64

A conducting spherical layer with inner radius $R_1$ and outer radius $R_2$ carries no net charge. A point charge $q$ sits inside the cavity, a distance $r < R_1$ from the center $O$ . Find the potential at $O$ .

The uncharged conducting layer, radii R_1 and R_2 about O, with the point charge q in the cavity a distance r from O. — The uncharged conducting layer, radii $R_1$ and $R_2$ about $O$ , with the point charge $q$ in the cavity a distance $r$ from $O$ .

Given

point charge $q$
$r < R_1$
layer radii $R_1$ , $R_2$
layer uncharged

Find

the potential

\varphi

O

Solution

Step 1How much charge each surface carries

In equilibrium the field inside the metal is zero. A closed surface drawn wholly inside the metal, around the cavity, therefore has zero flux, so the charge it encloses, $q$ plus the charge $q_1$ on the inner surface, is zero.

\oint \vect{E}\cdot d\vect{S} = \frac{q + q_1}{\varepsilon_0} = 0

The layer is uncharged, so the outer surface carries the opposite of the inner one.

q_1 = -q, \qquad q_2 = -q_1 = +q \tag{1}

Step 2How the charge is spread

The inner charge $-q$ crowds toward $q$ and its density is not uniform. The outer charge $+q$ is uniform: the metal screens the cavity, so the outer surface sees no field from $q$ and the inner charge together and spreads its charge as a lone charged sphere does.

The induced charge: -q on the inner surface, crowded toward q; +q spread evenly over the outer surface. The lines from q end on the inner surface and none cross the metal; outside, the field is that of +q at O. — The induced charge: $-q$ on the inner surface, crowded toward $q$ ; $+q$ spread evenly over the outer surface. The lines from $q$ end on the inner surface and none cross the metal; outside, the field is that of $+q$ at $O$ .

Step 3Superposing the three charges at

O

The potential at $O$ is the sum of the point charge’s contribution, a distance $r$ away, the inner surface’s, every element $dq_1$ a distance $R_1$ away, and the outer surface’s, every element $dq_2$ a distance $R_2$ away.

\begin{aligned} \varphi(O) &= \frac{1}{4\pi\varepsilon_0}\frac{q}{r} + \frac{1}{4\pi\varepsilon_0}\oint \frac{dq_1}{R_1} + \frac{1}{4\pi\varepsilon_0}\oint \frac{dq_2}{R_2}\\ &= \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r} + \frac{q_1}{R_1} + \frac{q_2}{R_2}\right) \end{aligned}

Substituting the totals (1):

\varphi = \dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{r} - \dfrac{1}{R_1} + \dfrac{1}{R_2}\right)

(2)

Check

Let the layer shrink to zero thickness, $R_2 \to R_1$ . The inner and outer charges then sit on the same sphere and cancel, and (2) becomes

\varphi \to \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r} - \frac{1}{R_1} + \frac{1}{R_1}\right) = \frac{q}{4\pi\varepsilon_0 r},

the potential of $q$ alone, as it must be with no conductor left. For $r < R_1 < R_2$ the bracket is positive, so $\varphi > 0$ for $q > 0$ .

Remark

Only the last term knows about the outside world. The outer surface charge is uniform wherever $q$ sits in the cavity, so the field outside the layer is that of $+q$ placed at $O$ : a closed conductor hides the position of a charge inside it, but not the charge itself. Earthing the layer would remove $q_2$ and with it the term $1/R_2$ , leaving $\varphi(O) = (q/4\pi\varepsilon_0)(1/r - 1/R_1)$ .