Skip to main content Contents Index Prev Up Next \(\newcommand{\vf}[1]{\mathbf{\boldsymbol{\vec{#1}}}}
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\)
Section 15.2 The Divergence of a Coulomb Field
The electric field of a point charge at the origin is given by
\begin{gather*}
\EE = \frac{1}{4\pi\epsilon_0} \frac{q\,\rhat}{r^2}
\end{gather*}
We can take the divergence of this field using the expression in
Section A.1 for the divergence of a radial vector field in spherical coordinates, which yields
\begin{gather*}
\grad\cdot\EE
= \frac{1}{r^2} \Partial{}{r}\Bigl(r^2 E_r\Bigr)
= \frac{1}{4\pi\epsilon_0} \Partial{q}{r}
= 0
\end{gather*}
On the other hand, the flux of this electric field through a sphere centered at the origin is
\begin{gather*}
\Int_{\textrm{sphere}} \EE\cdot d\AA
= \Int_{\textrm{sphere}} \!\!
\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \> dA
= \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \left( 4\pi r^2 \right)
= \frac{q}{\epsilon_0}
\end{gather*}
in agreement with Gauss’ Law. The Divergence Theorem then tells us that
\begin{gather*}
\Int_{\textrm{inside}} \grad\cdot\EE \> \dV
= \Int_{\textrm{sphere}} \!\! \EE\cdot d\AA
\ne 0
\end{gather*}
even though \(\grad\cdot\EE=0\text{!}\) What’s going on?
A bit of thought yields a clue: \(\EE\) isn’t defined at \(r=0\text{;}\) neither is its divergence. So we have a function which vanishes almost everywhere, whose (flux) integral isn’t zero. This should remind you of the Dirac delta function. However, we’re in 3 dimensions here, so that the correct conclusion is
\begin{gather*}
\grad\cdot\frac{1}{4\pi\epsilon_0} \frac{q\,\rhat}{r^2}
= \frac{q}{\epsilon_0} \, \delta^3(\rr)
= \frac{q}{\epsilon_0} \, \delta(x) \, \delta(y) \, \delta(z)
\end{gather*}
or equivalently
\begin{gather*}
\rho = q \, \delta^3(\rr)
\end{gather*}
which should not be surprising.
For a discussion of the delta function, please see
Chapter 8 .
