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[tex]\nabla \cdot \left(\frac{\hat{r}}{r^2}\right) = 4\pi \delta^3(\vec{r})[/tex]

Applying this result to Coulomb law, I get that

[tex]\nabla \cdot \vec{E} = \frac{q}{4 \pi \epsilon_0}4\pi \delta^3(\vec{r}) = \frac{q}{\epsilon_0}\delta^3(\vec{r})=\left\{ \begin{array}{rcl} q/\epsilon_0 & \mbox{for} & \vec{r} = \vec{0} \\ 0 & \mbox{elsewhere}

\end{array}[/tex]

Is this true? I'm wondering because Griffiths never makes mention of this result and I've never seen it anywhere else.

Note: Read [itex]\vec{r}[/itex] has the vector going from the point charge to some point P.