In David J. Griffiths's Introduction to Electrodynamics, the author gave the following problem in an exercise.
Sketch the vector function $$ \vec{v} ~=~ \frac{\hat{r}}{r^2}, $$ and compute its divergence, where $$\hat{r}~:=~ \frac{\vec{r}}{r} , \qquad r~:=~|\vec{r}|.$$ The answer may surprise you. Can you explain it?
I found the divergence of this function as $$ \frac{1}{x^2+y^2+z^2} $$ Please tell me what is the surprising thing here.
Answer
Pretty sure the question is about $\frac{\hat{r}}{r^2}$, i.e. the electric field around a point charge. Naively the divergence is zero, but properly taking into account the singularity at the origin gives a delta-distribution.
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