Irodov 3.63: Ask the Center, Not the Surface – Amit & Deepali Sinha Foundation

Problem Irodov 3.63

A conducting sphere carries no net charge. A point charge $q$ is placed outside it, at a distance $l$ from the sphere’s center. Find the potential $\varphi$ of the sphere.

The uncharged conducting sphere of radius R and the point charge q outside it, a distance l from the center. — The uncharged conducting sphere of radius $R$ and the point charge $q$ outside it, a distance $l$ from the center.

Given

point charge $q$
$l$ from the center
net charge of the sphere zero

Find

the sphere’s potential

\varphi

Solution

Step 1The sphere’s potential is the potential at its center

In equilibrium the whole conductor, interior included, is one equipotential. So the potential of the sphere equals the potential at any point inside it, and the center $O$ is the point where that potential is easiest to compute.

\varphi = \varphi(O) \tag{1}

Step 2Superposing the charge and the induced charge at the center

The potential at $O$ is the sum of the contribution of $q$ , a distance $l$ away, and that of the induced surface charge, each element $dq'$ of which is a distance $R$ away. The induced charge redistributes but its total is zero, because the sphere is uncharged.

The field of the solved problem: induced - charge on the side facing q, + charge on the far side, lines meeting the surface at right angles. Inside the sphere, the images q'_1 and q'_2 (dashed) reproduce that field outside. — The field of the solved problem: induced $-$ charge on the side facing $q$ , $+$ charge on the far side, lines meeting the surface at right angles. Inside the sphere, the images $q'_1$ and $q'_2$ (dashed) reproduce that field outside.

Writing the sum, with the surface integral running over the induced charge:

\begin{aligned} \varphi(O) &= \frac{1}{4\pi\varepsilon_0}\frac{q}{l} + \frac{1}{4\pi\varepsilon_0}\oint \frac{dq'}{R} \\ &= \frac{1}{4\pi\varepsilon_0}\frac{q}{l} + \frac{1}{4\pi\varepsilon_0 R}\oint dq' \\ &= \frac{1}{4\pi\varepsilon_0}\frac{q}{l} + \frac{0}{4\pi\varepsilon_0 R} \end{aligned}

By (1) this is the potential of the sphere.

\varphi = \dfrac{q}{4\pi\varepsilon_0 l}

(2)

Check

The image solution gives the same value. For a grounded sphere the image $q'_1 = -qR/l$ at distance $R^2/l$ from $O$ holds the whole surface at zero; to make the total induced charge zero, an image $q'_2 = +qR/l$ is added at the center. The first pair contributes nothing on the surface, so the surface potential is that of $q'_2$ alone, at distance $R$ :

\varphi = \frac{1}{4\pi\varepsilon_0}\frac{q'_2}{R} = \frac{1}{4\pi\varepsilon_0}\frac{qR/l}{R} = \frac{q}{4\pi\varepsilon_0 l}

in agreement with (2). As $l \to \infty$ the potential falls to zero, the value of an isolated uncharged sphere with nothing nearby.

Remark

The radius never enters: an uncharged sphere of any size, with $q$ at a distance $l$ from its center, sits at the potential that $q$ alone would produce at that center. Green’s reciprocity theorem generalizes this to a conductor of any shape: the potential an outside charge $q$ gives an uncharged conductor equals $q/Q$ times the potential that the conductor, carrying a charge $Q$ alone, produces at the place where $q$ sits. For a sphere that potential is $Q/(4\pi\varepsilon_0 l)$ :

\varphi = \frac{q}{Q}\,\frac{Q}{4\pi\varepsilon_0 l} = \frac{q}{4\pi\varepsilon_0 l}