# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

##### The Energy of a Continuous Charge Distribution
ems.es.e.ccd

For this case, \ref{Gr(2.42)} becomes

\[ W = \frac{1}{2} \int \rho({\bf r}) V({\bf r}) ~d\tau \label{Gr(2.43)} \]

Eliminate \(\rho\) and \(V\) in favor of \({\bf E}\):

\[ \rho = \varepsilon_0 {\boldsymbol \nabla} \cdot {\bf E} \longrightarrow W = \frac{\varepsilon_0}{2} \int ({\boldsymbol \nabla} \cdot {\bf E}) ~V d\tau. \]

Doing integration by parts and using \({\boldsymbol \nabla} V = -{\bf E}\),

\[ W = \frac{\varepsilon_0}{2} \left( \int_{\cal V} E^2 d\tau + \oint_{\cal S} V {\bf E} \cdot d{\bf a} \right) \label{Gr(2.44)} \]

Integrating over all space,

\[ W = \frac{\varepsilon_0}{2} \int E^2 d\tau \label{eq:Energy_as_int_E2} \]

Example 2.8: Find the energy of a uniformly charged spherical shell of total charge \(q\) and radius \(R\).

Solution 1: use \ref{Gr(2.43)} in version for surface charges:

\[ W = \frac{1}{2} \int \sigma V da. \]

The potential at the surface of the sphere is constant, \(V = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}\), so

\[ W = \frac{1}{8\pi \varepsilon_0} \frac{q}{R} \int \sigma da = \frac{1}{8\pi \varepsilon_0} \frac{q^2}{R}. \]

Solution 2: use \ref{eq:Energy_as_int_E2}. Inside sphere: \({\bf E} = 0\); outside,

\[ {\bf E} = \frac{1}{4\pi \varepsilon_0} q \frac{\hat{\bf r}}{r^2} \rightarrow E^2 = \frac{q^2}{(4\pi \varepsilon_0)^2 r^4}. \]

so \[ W = \frac{\varepsilon_0}{2} \int_{outside~sphere} \frac{q^2}{(4\pi \varepsilon_0)^2 r^4} r^2 \sin \theta dr d\theta d\phi = \frac{1}{32 \pi^2 \varepsilon_0} q^2 4\pi \int_R^{\infty} \frac{dr}{r^2} = \frac{1}{8\pi \varepsilon_0} \frac{q^2}{R}. \]

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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