Divergence of inverse square field (2D vs. 3D)

Question

Let $\displaystyle\mathbf{v}=\frac{\mathbf{\hat{r}}}{r^2}$.

Compute its divergence.

My attempt:

I found that $\nabla\cdot\mathbf{v}=0$ in the 3D case, in accordance with Gauss' law, but $\nabla\cdot\mathbf{v}\ne 0$ in the 2D case. Is this correct or am I missing something?

Answer 1

It is correct. Try to calculate the electric field due to an infinitely long straight wire. You will see that the field is proportional to $1/r$, and not $1/r^2$ like in 3D. So in 2D is just "inverse distance" not "inverse squared distance".