Skip to main content

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.