Irodov 3.130: Charged Balls Interact Like Points – Amit & Deepali Sinha Foundation

Problem Irodov 3.130

Two balls carry charges $q_1$ and $q_2$ , each distributed with spherical symmetry about its ball’s center. The centers are a distance $l$ apart. Find the interaction energy of the two balls. (Hint: first find the interaction energy of a ball and a thin spherical layer of charge.)

Two balls with spherically symmetric charges q_1 and q_2, centers O_1 and O_2 a distance l apart. — Two balls with spherically symmetric charges $q_1$ and $q_2$ , centers $O_1$ and $O_2$ a distance $l$ apart.

Given

charges $q_1$ , $q_2$ , each spherically symmetric
distance $l$ between centers
the balls do not overlap (assumed)

Find

the interaction energy

W

Solution

Step 1The field outside a spherically symmetric charge

We take the balls not to overlap, so each lies entirely outside the other. Outside a spherically symmetric charge the field is radial and the same all round, so Gauss’s theorem over a concentric sphere gives the field, and hence the potential, of a point charge at the center. For ball 1 at a distance $s$ from $O_1$ beyond its surface:

\begin{aligned} E\cdot 4\pi s^2 &= \frac{q_1}{\varepsilon_0}\\ E &= \frac{q_1}{4\pi\varepsilon_0 s^2}\\ \varphi_1(s) &= \int_s^\infty E(s')\,ds' = \frac{q_1}{4\pi\varepsilon_0 s} \end{aligned} \tag{1}

The same holds for a thin uniform spherical layer at every point outside it.

Step 2A ball and one thin spherical layer

Split ball 2 into thin concentric layers. One of them, of radius $r$ about $O_2$ , carries the charge $dq_2$ , spread uniformly over it by the spherical symmetry.

Ball 1 and one thin layer of ball 2: radius r about O_2, charge dq_2 (drawn positive), lying wholly outside ball 1. — Ball 1 and one thin layer of ball 2: radius $r$ about $O_2$ , charge $dq_2$ (drawn positive), lying wholly outside ball 1.

The interaction energy of the layer with ball 1 is the work to bring the charge of ball 1 in from infinity against the layer’s field alone, so it is the potential of the layer summed over the charge of ball 1:

dW = \int_{\text{ball 1}}\varphi_{\text{layer}}\,dq_1

Every point of ball 1 lies outside the layer, where by the remark under (1) the layer acts as the point charge $dq_2$ at $O_2$ . Substituting that potential, the remaining integral over ball 1 is the potential of ball 1 at $O_2$ , given by (1) at the distance $l$ from $O_1$ :

\begin{aligned} dW &= \int_{\text{ball 1}}\frac{dq_2\,dq_1}{4\pi\varepsilon_0\lvert\vect r_1 - \vect r_{O_2}\rvert} = dq_2\,\varphi_1(l)\\ &= \frac{q_1\,dq_2}{4\pi\varepsilon_0 l} \end{aligned} \tag{2}

Step 3Summing the layers of ball 2

The result (2) does not depend on the radius $r$ of the layer, so adding the layers of ball 2 simply adds their charges to $q_2$ :

W = \int dW = \frac{q_1}{4\pi\varepsilon_0 l}\int dq_2

W = \frac{q_1q_2}{4\pi\varepsilon_0 l}

(3)

Check

Dimensions: $q_1q_2/(\varepsilon_0 l)$ is an energy. Limit: shrinking both balls to points leaves the formula unchanged, and it is then exactly the interaction energy of two point charges, as it must be. Force: $-dW/dl = q_1q_2/(4\pi\varepsilon_0 l^2)$ along the line of centers, the Coulomb force between the centers, which is also what the outside field (1) of one ball exerts on the other’s net charge.

Remark

The result breaks down the moment the balls overlap: the charge of ball 1 that lies inside a layer of ball 2 sits in that layer’s constant inner potential, not in a point-charge potential, and the energy falls below $q_1q_2/(4\pi\varepsilon_0 l)$ for like charges. This is the electrostatic twin of Newton’s shell theorem, which lets two planets of any spherically layered structure attract exactly as point masses for as long as they do not touch.