Irodov 3.135: One Sixth of the Energy Inside – Amit & Deepali Sinha Foundation

Problem Irodov 3.135

A ball of radius $R$ carries a charge $q$ spread uniformly through its volume; the permittivity is $1$ everywhere. Find:

the electrostatic self energy of the ball;
the ratio of the field energy $W_1$ inside the ball to the field energy $W_2$ in the space outside it.

The ball of radius R about O, with the charge q spread uniformly through its volume (drawn for q > 0). — The ball of radius $R$ about $O$ , with the charge $q$ spread uniformly through its volume (drawn for $q > 0$ ).

Given

charge $q$ uniform through a ball of radius $R$
permittivity $\varepsilon = 1$

Find

(a) the self energy

W

; (b) the ratio

W_1/W_2

Solution

Step 1The field inside and outside

The charge is spherically symmetric, so the field is radial and the same all over a concentric sphere of radius $r$ . Gauss’s theorem over that sphere encloses the fraction $(r/R)^3$ of the charge inside the ball and all of it outside:

E = \begin{cases} \dfrac{q\,r}{4\pi\varepsilon_0 R^3}, & r < R\\[2ex] \dfrac{q}{4\pi\varepsilon_0 r^2}, & r > R \end{cases} \tag{1}

The self energy of the ball is the energy of this field, with density $w = \varepsilon_0E^2/2$ , summed over thin spherical layers of volume $4\pi r^2\,dr$ .

Step 2The energy inside the ball

Substituting the inner field of (1) into the density and integrating from the center to the surface:

\begin{aligned} W_1 &= \int_0^R\frac{\varepsilon_0}{2}\left(\frac{q\,r}{4\pi\varepsilon_0R^3}\right)^2 4\pi r^2\,dr = \frac{q^2}{8\pi\varepsilon_0R^6}\int_0^R r^4\,dr \end{aligned}

W_1 = \frac{q^2}{40\pi\varepsilon_0 R}

(2)

Step 3The energy outside the ball

With the outer field of (1), from the surface to infinity:

W_2 = \int_R^\infty\frac{\varepsilon_0}{2}\left(\frac{q}{4\pi\varepsilon_0r^2}\right)^2 4\pi r^2\,dr = \frac{q^2}{8\pi\varepsilon_0}\int_R^\infty\frac{dr}{r^2}

W_2 = \frac{q^2}{8\pi\varepsilon_0 R}

(3)

Step 4(a) The self energy

The self energy is the whole field energy, the sum of (2) and (3), with $1/8 = 5/40$ :

W = \frac{q^2}{40\pi\varepsilon_0R} + \frac{5q^2}{40\pi\varepsilon_0R} = \frac{6q^2}{40\pi\varepsilon_0R}

W = \frac{3q^2}{20\pi\varepsilon_0 R}

(4)

Step 5(b) Where the energy sits

Dividing (2) by (3), the factor $q^2/(8\pi\varepsilon_0 R)$ cancels:

\frac{W_1}{W_2} = \frac{1}{5}

(5)

Only one sixth of the self energy is stored inside the ball. The integrands above show why: the energy per unit radius grows as $r^4$ inside and falls only as $1/r^2$ outside.

The field energy per unit radius, dW/dr = 4π r^2 w: it rises as r^4 inside the ball, peaks at the surface, and falls as 1/r^2 outside, where most of the energy lies. — The field energy per unit radius, $dW/dr = 4\pi r^2 w$ : it rises as $r^4$ inside the ball, peaks at the surface, and falls as $1/r^2$ outside, where most of the energy lies.

Check

Dimensions: $q^2/(\varepsilon_0 R)$ is an energy. Independent route through the work of assembly: building the ball layer by layer, a ball of radius $r$ carries the charge $q(r/R)^3$ , so its surface potential is $q r^2/(4\pi\varepsilon_0R^3)$ , and the next layer brings the charge $dq = 3q\,r^2dr/R^3$ in from infinity to that potential:

\begin{aligned} W &= \int_0^R\frac{q r^2}{4\pi\varepsilon_0R^3}\cdot\frac{3q\,r^2}{R^3}\,dr\\ &= \frac{3q^2}{4\pi\varepsilon_0R^6}\cdot\frac{R^5}{5}, \end{aligned}

which is the boxed self energy.

Remark

Setting this energy equal to $m_ec^2$ for an electron pictured as a uniform ball gives the radius $R = 3e^2/(20\pi\varepsilon_0 m_ec^2) \approx 1.7\times10^{-15}$ m, the size of the classical electron radius. It is not the electron’s real size, which experiments show to be far smaller, but it marks the scale below which classical electrostatics of a point charge stops making sense.