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+### Statement
+
+$11.2.21.$ [Insert the problem statement]
+
+### 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:
+
+$m_{\text{em}} = \frac{U}{c^2} = \frac{1.152 \times 10^{-10}}{(3.0 \times 10^8)^2} = \frac{1.152 \times 10^{-10}}{9.0 \times 10^{16}} = 1.28 \times 10^{-27} \, \text{kg}$
+
+Result
+
+$\boxed{m_{\text{em}} \approx 1.3 \times 10^{-27} \text{ kg}}$
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]