Problem 11.2.21^∗

Solution

We model the electron as a sphere of radius R with charge e distributed on its surface. The electrostatic potential energy in SI is:

$U = \frac{1}{2} \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{R}$

The equivalent electromagnetic mass, according to $E = mc^2$ is

$m_{\text{em}} = \frac{U}{c^2}$

Data

Electron charge: $e = 1.60 \times 10^{-19} \text{ C}$
Coulomb constant: $\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2$

Radius (experimental limit): $R = 10^{-18} \text{ m}$

Speed of light: $c = 3.0 \times 10^8 \text{ m/s}$

Calculation

$e^2 = (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \frac{1}{4\pi\varepsilon_0} e^2 = (9.0 \times 10^9) \times (2.56 \times 10^{-38}) = 2.304 \times 10^{-28} \, \text{J·m}$

Potential energy U:

$U = \frac{1}{2} \times \frac{2.304 \times 10^{-28}}{10^{-18}} = \frac{1}{2} \times 2.304 \times 10^{-10} = 1.152 \times 10^{-10} \, \text{J}$

Electromagnetic mass:

$m_{\text{em}} = \frac{U}{c^2} = \frac{1.152 \times 10^{-10}}{(3.0 \times 10^8)^2} = \frac{1.152 \times 10^{-10}}{9.0 \times 10^{16}} = 1.28 \times 10^{-27} \, \text{kg}$