Why the relation $\mathbf{E}=-\nabla \psi$ holds at points $P \in V'$?

Question

Let there be a continuous charge distribution in space having volume $V'$ and density $\rho$.

Let:

$\displaystyle \mathbf{E}=\int_{V'} \rho\ \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV'$

$\displaystyle \psi=\int_{V'} \dfrac{\rho}{|\mathbf{r}-\mathbf{r'}|} dV'$

The existence of $\mathbf{E}$ and $\psi$ at a point $P \in V'$ can be shown by switching to spherical coordinate system.

However it is not obvious to me why the relation $\mathbf{E}=-\nabla \psi$ holds at points $P \in V'$. Can anybody please show this to me?

Answer 1

We have that (assuming that we can interchange the order of differentiation and integration) \begin{align} \nabla \psi(r) &=\nabla_r\int_{V'} \frac{\varrho(r')}{|r-r'|}\, \mathrm{d}^3r'\\ &=\int_{V'}\varrho(r')\nabla_r\frac{1}{|r-r'|}\, \mathrm{d}^3r'\\ &=-\int_{V'}\varrho(r')\frac{r-r'}{|r-r'|^3}\, \mathrm{d}^3r'\\ \end{align} Because we have that \begin{align} (\nabla f(|r|))_i &=\partial_i f(|r|)\\ &=f'(|r|) \partial_i |r|\\ &= f'(|r|)\partial_i \sqrt{x_j x_j}\\ &= f'(|r|) \frac{1}{2\sqrt{x_jx_j}} \partial_i(x_kx_k)\\ &= f'(|r|) \frac{1}{\sqrt{x_jx_j}} x_i\\ &= f'(|r|) \frac{x_i}{|x|} \end{align} And if we let $f(x)=x^{-1}$ we have that $f'(x)=-x^{-2}$, as we wanted.