Irodov 3.134: Two Shells, Three Energies – Amit & Deepali Sinha Foundation

Problem Irodov 3.134

Two thin concentric metal shells have radii $R_1$ and $R_2$ and carry charges $q_1$ and $q_2$ respectively. Find the self energy $W_1$ of the first shell, the self energy $W_2$ of the second, the interaction energy $W_{12}$ of the two shells, and the total electric energy of the system.

The two concentric shells about O: the inner of radius R_1 with charge q_1, the outer of radius R_2 with charge q_2 (both drawn positive). — The two concentric shells about $O$ : the inner of radius $R_1$ with charge $q_1$ , the outer of radius $R_2$ with charge $q_2$ (both drawn positive).

Given

thin concentric shells, radii $R_1$ , $R_2$
charges $q_1$ , $q_2$
$R_1 < R_2$ (assumed)

Find

W_1

W_2

W_{12}

and the total energy

W

Solution

Step 1The two fields and the energy they share

Take the first shell to be the inner one, $R_1 < R_2$ . Each shell’s charge spreads uniformly over it, so by symmetry its field is radial and the same all round: zero inside the shell, and outside it, by Gauss’s theorem $E\cdot 4\pi r^2 = q/\varepsilon_0$ ,

\begin{aligned} E_1 &= \frac{q_1}{4\pi\varepsilon_0 r^2} \quad (r > R_1)\\ E_2 &= \frac{q_2}{4\pi\varepsilon_0 r^2} \quad (r > R_2) \end{aligned} \tag{1}

and zero at smaller $r$ . The field of the system is $\vect E_1 + \vect E_2$ , and squaring it in the energy density splits the energy into three parts:

\begin{aligned} W &= \int\frac{\varepsilon_0 E_1^2}{2}\,dV + \int\frac{\varepsilon_0 E_2^2}{2}\,dV + \int\varepsilon_0\,\vect E_1\cdot\vect E_2\,dV\\ &= W_1 + W_2 + W_{12} \end{aligned} \tag{2}

Step 2The self energy of each shell

For a shell of charge $q$ and radius $R$ , the field of (1) fills the space outside it; integrating the energy density over spherical layers $dV = 4\pi r^2 dr$ :

\begin{aligned} \int_R^\infty\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_R^\infty\frac{dr}{r^2}\\ &= \frac{q^2}{8\pi\varepsilon_0 R} \end{aligned}

With $q_1, R_1$ and $q_2, R_2$ in turn:

\begin{gathered} W_1 = \frac{q_1^2}{8\pi\varepsilon_0 R_1}\\ W_2 = \frac{q_2^2}{8\pi\varepsilon_0 R_2} \end{gathered}

(3)

Step 3The interaction energy

The cross term of (2) is nonzero only where both fields exist, outside the outer shell, $r > R_2$ ; there the two fields of (1) are parallel radial vectors:

\begin{aligned} W_{12} &= \int_{R_2}^\infty\varepsilon_0\,\frac{q_1}{4\pi\varepsilon_0 r^2}\,\frac{q_2}{4\pi\varepsilon_0 r^2}\,4\pi r^2\,dr\\ &= \frac{q_1q_2}{4\pi\varepsilon_0}\int_{R_2}^\infty\frac{dr}{r^2} \end{aligned}

W_{12} = \frac{q_1q_2}{4\pi\varepsilon_0 R_2}

(4)

It depends on the outer radius only: the inner shell’s potential on the outer shell is $q_1/(4\pi\varepsilon_0 R_2)$ , and $W_{12}$ is $q_2$ times it.

Step 4The total energy

Adding the three parts, as (2) requires:

W = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1^2}{2R_1} + \frac{q_2^2}{2R_2} + \frac{q_1q_2}{R_2}\right)

(5)

Check

Through the potentials of the shells, $W = (q_1\varphi_1 + q_2\varphi_2)/2$ , with the inner shell at $\varphi_1 = (q_1/R_1 + q_2/R_2)/(4\pi\varepsilon_0)$ (the outer shell’s potential is constant inside it) and the outer at $\varphi_2 = (q_1 + q_2)/(4\pi\varepsilon_0 R_2)$ :

W = \frac{1}{8\pi\varepsilon_0}\left(\frac{q_1^2}{R_1} + \frac{q_1q_2}{R_2} + \frac{q_1q_2}{R_2} + \frac{q_2^2}{R_2}\right),

the same as the boxed total. For $q_2 = -q_1$ it reduces to $q_1^2(1/R_1 - 1/R_2)/(8\pi\varepsilon_0)$ , the energy $q^2/2C$ of a spherical capacitor. Had the first shell been the outer one, $R_1 > R_2$ , the cross term would take $R_1$ : in general $W_{12} = q_1q_2/(4\pi\varepsilon_0 R_{\text{outer}})$ .

Remark

Self energies are always positive, while $W_{12}$ takes the sign of $q_1q_2$ and can make the total smaller than the sum of the self energies. In the capacitor case it is even smaller than the inner shell’s self energy alone, because the field outside the outer shell cancels completely. Shrinking a shell to a point sends its self energy to infinity; that constant is why the interaction energy alone is the useful quantity for point charges.