##### Example calculations for the potential
ems.es.ep.ex

Spherical shell: via \({\bf E}\)

Consider a spherical shell of radius \(R\) with uniform surface charge density and total charge \(q\) (so \(\sigma = q/4\pi R^2\)). The shell is centered at the origin.

We want to know the potential throughout space. Starting from the field as obtained from applying Gauss's law Gl_i,

\[ {\bf E} = \frac{1}{4\pi \varepsilon_0} q \frac{\hat{\bf r}}{r^2}, \hspace{5mm} r > R, \hspace{2cm} {\bf E} = 0, \hspace{5mm} r < R. \]

Adopting our usual convention of putting the reference point at infinity, and using spherical symmetry, we get:

- for \(r > R\):

\[ \phi(r > R) = -\int_{{\cal O}: \infty}^{\bf r} {\bf E} \cdot d{\bf l} = -\frac{1}{4\pi \varepsilon_0} \int_{\infty}^r q \frac{dr'}{r'^2} = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}, \]

- for \(r < R\): the integrand vanishes for the part \(r < R\), and

\[ \phi(r < R) = \phi(R) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}. \]

Spherical shell: direct calculation

We can also compute the potential for the uniformly-charged shell using a direct integral, starting from p_scd

\[ \phi({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int da' \frac{\sigma}{|{\bf r} - {\bf r}'|}. \]

Exploiting spherical symmetry, let us orient \({\bf r}\) along the \(\hat{\bf z}\) axis so \({\bf r} = z \hat{\bf z}\). Our integration point \({\bf r}'\) is at position \(R \sin \theta' \cos \varphi' \hat{{\bf x}} + R \sin \theta' \sin \varphi' \hat{\bf y} + R\cos \theta' \hat{\bf z}\) so

\begin{equation*} |{\bf r} - {\bf r}'|^2 = R^2 \sin^2 \theta' + (R \cos \theta' - z)^2 = R^2 + z^2 - 2Rz \cos \theta' \end{equation*}

Using the element of surface area (in spherical coordinates) \(R^2 \sin \theta' d\theta' d\varphi'\), so

\begin{align} 4\pi \varepsilon_0 \phi(z) &= \sigma \int_0^{\pi} d\theta' \int_0^{2\pi} d\varphi' \frac{R^2 \sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\ &= 2\pi R^2 \sigma \int_0^{\pi} d\theta' \frac{\sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\ &= 2\pi R^2 \sigma \left. \left( \frac{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}}{Rz} \right)\right|_0^{\pi} \nonumber \\ &= \frac{2\pi R\sigma}{z} \left(\sqrt{R^2 + z^2 + 2Rz} - \sqrt{R^2 + z^2 - 2Rz} \right) \nonumber \\ &= \frac{2\pi R \sigma}{z} \left( \sqrt{(R + z)^2} - \sqrt{(R - z)^2} \right). \end{align}

We must obviously choose the positive root \(\sqrt{(R + z)^2} = R+z\) and thus

\begin{align} \phi(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (z-R)) = \frac{R^2 \sigma}{\varepsilon_0 z}, \hspace{1cm} \mbox{outside}, \nonumber \\ \phi(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (R-z)) = \frac{R \sigma}{\varepsilon_0}, \hspace{1cm} \mbox{inside}. \end{align}

In terms of total charge on the shell, \(q = 4\pi R^2 \sigma\), we thus obtain \(\phi(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{z}\) outside, and \(\phi(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}\) inside which indeed correspond to our previous solution via \({\bf E}\).

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42