#### Electrostatic Energy from the Potential
ems.es.e

- FLS II 8
- Gr 2.4
- PM 2

We have already seen that the electrostatic potential is the potential energy of a unit charge brought from infinity, and that since the force is proportional to the charge we're moving, we can simply calculate the work per unit charge \(W_u\) (so \(W = q W_u\)):

\[ W_u = -\int_{{\bf a}}^{{\bf b}} {\bf E} \cdot d{\bf l} \]

For a point charge distribution, the total work required to assemble it can also be rewritten in terms of the potential by using p,

- Gr (2.42)

\[ W = \frac{1}{2} \sum_{i=1}^m q_i \phi({\bf r}_i) \tag{W_pcd}\label{W_pcd} \]

For a volume charge distribution, this becomes

- Gr (2.43)

\[ W = \frac{1}{2} \int \rho({\bf r}) \phi({\bf r}) ~d\tau \tag{W_vcd}\label{W_vcd} \]

In this, we can eliminate \(\rho\) and \(\phi\) in favour of \({\bf E}\):

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

Integrating by parts and using \({\boldsymbol \nabla} \phi = -{\bf E}\),

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

Integrating over all space (and thus dropping the surface integrale whose integrand is assumed to vanish at infinity), we get

- Gr (2.45)

\[ W = \frac{\varepsilon_0}{2} \int E^2 d\tau \tag{W_intEsq}\label{W_intEsq} \]

Energy of spherical shell

Consider a uniformly charged spherical shell of total charge \(q\) and radius \(R\). We wish to compute its electrostatic energy.

- First way: we use W_vcd adapted for surface charges,

\[ W = \frac{1}{2} \int_S da ~\sigma \phi. \]

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

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

- Second way: we use W_intEsq. Inside the 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

\begin{align} 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\varphi \nonumber \\ &= \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}. \end{align}

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42