Self-Energy of a Uniformly Charged Spherical Surface: 04-BS-9, December 2014, Q7

Question 7: Self-Energy of a Uniformly Charged Spherical Surface (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Total charge $Q$	$e=1.6\times10^{-19}$ C, uniform on the surface
Sphere radius $R$	$10^{-10}$ m

Find. The total electrostatic (self-assembly) energy of the charged surface.

Approach. Use the standard result for the self-energy of a uniformly charged spherical surface, obtained by integrating the work to bring successive charge elements from infinity onto the sphere (equivalently, integrating the field energy density $\varepsilon_0E^2/2$ over all space outside the sphere, where $E=0$ inside). This is the same self-energy calculation used to bound the classical electron radius, and it is worth noting up front that $R=10^{-10}$ m here is atomic-scale (roughly a Bohr-radius-sized layer of charge), not nuclear-scale, so the resulting energy should land in the eV range rather than the MeV range typical of nuclear self-energies.

Self-energy of a charged spherical surface. $$U=\frac{kQ^2}{2R}=\frac{(8.992\times10^9)(1.6\times10^{-19})^2}{2(10^{-10})}$$ $$U=\boxed{1.151\times10^{-18}\ \text{J}}$$
Convert to electron-volts for intuition. $$U_{\text{eV}}=\frac{U}{e}=\frac{1.151\times10^{-18}}{1.6\times10^{-19}}$$ $$U=\boxed{7.19\ \text{eV}}$$

Quantity	Result
$U$	$1.151\times10^{-18}$ J
$U$ (eV)	$7.19$ eV

Concept & Theory

Topic. Electrostatic self-energy of a uniformly charged spherical surface (thin conducting-style surface layer).

Key relations. $U=kQ^2/2R$ for a charged spherical surface (compare $U=3kQ^2/5R$ for a uniformly charged solid sphere — a different, larger, result because the interior also carries field energy in that case).

Why this works. Because the field inside a uniformly charged spherical surface is exactly zero (Gauss's law) and equals the point-charge field outside, the field-energy integral $\int(\varepsilon_0E^2/2)\,dV$ runs only from $R$ to infinity, giving exactly half the coefficient that a full point-charge self-energy calculation (which formally diverges) would suggest at $r=R$.

Common pitfall. Confusing the spherical-surface result ($kQ^2/2R$) with the uniformly-charged solid-sphere result ($3kQ^2/5R$) — the question specifies charge spread on the surface.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Energy in an electrostatic field, self-energy of charge distributions).

Question 7 of 8: Self-Energy of a Uniformly Charged Spherical Surface

Question 7: Self-Energy of a Uniformly Charged Spherical Surface (20 marks)