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For divergence of Coulomb field, we can directly apply the formula of divergence in spherical coordinate to show that $$\nabla\cdot\left(\frac{\hat{\mathbf{r}}}{r^2}\right)=0$$ However, the formula doesn't work for $r=0$. But we can consider a small volume enclosing the origin and show that the divergence is infinite at the origin, and that in fact it is a Dirac delta function.

What about curl?

We can similarly apply the formula of curl in spherical coordinate to show that $$\nabla\times\left(\frac{\hat{\mathbf{r}}}{r^2}\right)=0$$ at all $r\ne 0$.

But how to show that the curl is zero at the origin as well?

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  • $\begingroup$ You get $4\pi \delta^3(r)$ $\endgroup$ – Peter Diehr Aug 21 '16 at 1:28
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Curl is a vector. So, for a coulomb field around a point charge, given that there is NO preferred direction, one cannot believe a nonzero curl.

One definition of curl is a limit of a line integral, and with that definition, it can be computed at a point discontinuity (by enclosing the point with a closed curve, and taking the limit as the curve is reduced toward zero area). In particular, concentric circles around a point charge will converge on zero curl for the electric field. Heck, on concentric circles around a point charge, even the infinitesimal elements of the line integral are all individually zero.

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