Electric field and Potential

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Coulomb's law

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

Gauss law

[math]\oint_S \textbf{E}\cdot d\textbf{a} = \frac{Q_{enc}}{\epsilon_0}[/math]

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

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

which leads to

[math]\nabla\cdot\textbf{E} = \frac{\rho}{\epsilon_0}[/math]

Consequence of being central force

Integration of E field along a closed loop vanishes.

[math]\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[/math]

Therefore, applying Stokes' theorem to above equation yields

[math]\oint_C \textbf{E}\cdot d\textbf{l} = 0 = \int_S (\nabla\times\textbf{E})\cdot d\textbf{a}[/math]

which means

[math]\nabla\times\textbf{E} = 0[/math]

Potential

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

[math]\mathbf{E} = -\nabla V[/math]

If we plug this definition into Gauss law, we have

[math]\nabla^2 V = -\frac{\rho}{\epsilon_0}[/math]