Given the map $E: \mathbb{R}^3 \times \mathbb{R} \rightarrow \mathbb{R}^3$, such that (notice that this excercise is taken from Physics(Electrodynamics)). $$E(r,t) = -\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \theta(vt-r)e_r$$ (Theta is the Heaviside step-functions)
I want to find out what $div_r(E)(r,t)$ and $\frac{\partial E}{\partial t}$ is?
My results were that $\frac{\partial E}{\partial t} = - \frac{qv}{4 \pi \varepsilon r^2} \delta(vt-r)e_r$ (Delta is the Dirac-delta function). $div_r(E)(r,t)= \frac{q}{4 \pi \varepsilon_0}\frac{q}{r^2}e_r\delta(vt-r)-\frac{q}{\varepsilon_0}\delta(r)\theta(vt-r).$
Can anybody check this?
First of all, $\partial E/\partial t$ needs to be a vector field again. I don't know where $\varepsilon_0$ or the chain rule went here, either.
If you have a radial vector field $E = F(r)e_r$, isn't the standard formula $\operatorname{div}E = \dfrac 1{r^2}\dfrac{d}{dr}(r^2F(r))$? So please double-check your answer.