Electric field and Potential

Coulomb's law

$\mathbf{E}(\mathbf{r}) = \frac1{4\pi\epsilon_0}\int \frac{\rho(\mathbf{r}')\,d\tau'}{R^2}\hat{R}$

Gauss law

$\oint_S \textbf{E}\cdot d\textbf{a} = \frac{Q_{enc}}{\epsilon_0}$

Applying divergence theorem, and using $Q_{enc}=\int_V \rho(\mathbf{r}')d\tau'$, we have

$\int_V \nabla\cdot\textbf{E}d\tau' = \frac{1}{\epsilon_0}\int_V \rho(\mathbf{r}')d\tau'$

$\nabla\cdot\textbf{E} = \frac{\rho}{\epsilon_0}$

Consequence of being central force

Integration of E field along a closed loop vanishes.

$\oint_C \textbf{E}\cdot d\textbf{l} \propto \oint_C \frac{\hat{r}\cdot d\textbf{l}}{r^2} = \oint_C \frac{dr}{r^2} = \left.-\frac{1}{r}\right|_{p_1}^{p_1}=0$

Therefore, applying Stokes' theorem to above equation yields

$\oint_C \textbf{E}\cdot d\textbf{l} = 0 = \int_S (\nabla\times\textbf{E})\cdot d\textbf{a}$

which means

$\nabla\times\textbf{E} = 0$

Potential

As $\nabla\times\textbf{E} = 0$, E field can be expressed as a gradient of a scalar field. Let us define potential V such that

$\mathbf{E} = -\nabla V$

If we plug this definition into Gauss law, we have

$\nabla^2 V = -\frac{\rho}{\epsilon_0}$