这是一份 Imperial帝国理工大学 PHYS97046作业代写的成功案例
using the product and chain rules of differentiation in conjunction. In particular, if $f(r)=r^{n-1}$,
$$
\nabla \cdot \mathbf{r} r^{n-1}=\nabla \cdot\left(\hat{\mathbf{r}} r^{n}\right)=3 r^{n-1}+(n-1) r^{n-1}=(n+2) r^{n-1}
$$
This divergence vanishes for $n=-2$, except at $r=0$ (where $\hat{\mathbf{r}} / r^{2}$ is singular). This is relevant for the Coulomb potential
$$
V(r)=A_{0}=-\frac{q}{4 \pi \epsilon_{0} r}
$$
with the electric field
$$
\mathbf{E}=-\nabla V=\frac{q \hat{\mathbf{r}}}{4 \pi \epsilon_{0} r^{2}}
$$
we obtain the divergence $\nabla \cdot \mathbf{E}=0$ (except at $r=0$, where the derivatives are undefined).
PHYS97046 COURSE NOTES ：
Suppose we eliminate B . We may do this by taking the curl of both sides of and the time derivative of both sides. Since the space and time derivatives commute,
$$
\frac{\partial}{\partial t} \nabla \times \mathbf{B}=\nabla \times \frac{\partial \mathbf{B}}{\partial t}
$$
and we obtain
$$
\nabla \times(\nabla \times \mathbf{E})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}
$$
Applicatio yields
$$
(\nabla \cdot \nabla) \mathbf{E}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}},
$$