Pre-Quantum Electrodynamics

Example 2.6: find the potential inside and outside a spherical shell or radius \(R\) centered at origin, which carries uniform surface charge. Ref point at infinity. Solution: from Gauss's law, \[ {\bf E} = \frac{1}{4\pi \varepsilon_0} q \frac{\hat{\bf r}}{r^2}, \hspace{5mm} r > R, \hspace{2cm} {\bf E} = 0, \hspace{5mm} r < R. \] Thus: using spherical symmetry, for \(r > R\): \[ V(r > R) = -\int_{{\cal O}: \infty}^{\bf r} {\bf E} \cdot d{\bf l} = -\frac{1}{4\pi \varepsilon_0} \int_{\infty}^r q \frac{dr'}{r'^2} = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}, \hspace{5mm} r > R. \] For \(r < R\), the integrand vanishes for the part \(r < R\), and \[ V(r < R) = V(R) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}. \]

Example 2.7: find the potential of a uniformly charged spherical shell of radius \(R\) (same problem as 2.6, but now done with \ref{Gr(2.30)}). Solution: Use \[ V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int da' \frac{\sigma}{|{\bf r} - {\bf r}'|}. \] Put \({\bf r}\) on \(\hat{\bf z}\) axis so \({\bf r} = z \hat{\bf z}\), use law of cosines:

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

Element of surface area: \(R^2 \sin \theta' d\theta' d\phi'\), so

\begin{align} 4\pi \varepsilon_0 V(z) &= \sigma \int_0^{\pi} d\theta' \int_0^{2\pi} d\phi' \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}

Take positive root:

\begin{align} V(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (z-R)) = \frac{R^2 \sigma}{\varepsilon_0 z}, \hspace{1cm} \mbox{outside}, \nonumber \\ V(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 shell, \(q = 4\pi R^2 \sigma\), \(V(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{z}\) outside, and \(V(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}\) inside.