Pre-Quantum Electrodynamics

Try to calculate this directly:

\begin{equation} {\boldsymbol \nabla} \cdot \frac{\hat{\bf r}}{r^2} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r^2}\right) =^{?} 0 \label{Gr(1.84)} \end{equation}

Now apply the divergence theorem. Integrate over a sphere of radius \(R\) centered at the origin (Prob. 1.38b):

\begin{equation} \oint \frac{\hat{\bf r}}{r^2} \cdot d{\bf a} = \int \left(\frac{1}{R^2} \hat{\bf r} \right) \cdot \left( R^2 \sin \theta d\theta d\varphi ~\hat{\bf r} \right) = \int_0^{\pi} d\theta \sin \theta \int_0^{2\pi} d\varphi = 4\pi \label{Gr(1.85)} \end{equation}

Problem: in Gr(1.84), we've divided by zero when \(r = 0\).