Gauss’s Law


\(\oint_S\vec{E}\cdot\hat{n}da = \frac{Q}{\varepsilon_0}\). \(\hat{n}\) pointing out of the surface (positively oriented).

Verifying with Coulomb’s Law

Single Point Charge

\(\frac{q}{4\pi\varepsilon_0}\oint_S\frac{\hat{r}\cdot\hat{n}}{r^2}dr = \frac{q}{4\pi\varepsilon_0}\oint_S\frac{\cos\theta dr}{r^2} = \frac{q}{4\pi\varepsilon_0}\oint_S\frac{da_\perp}{r^2} = \frac{q}{4\pi\varepsilon_0}\oint_S d\Omega = \frac{q}{\varepsilon_0}\) Note, this surface is closed so the integral resolves to the total solid angle.

Note, \(\ell = r\theta\) so \(A=r^2\Omega\)

Two Point Charges

Using linearity of Electric field, we see that Coulomb’s Law proves Gauss’s Law.

Verifying with Divergence Theorem

\(\oint_S\vec{E}\cdot\vec{n}da = \int_V\nabla\cdot\vec{E}dV = \frac{1}{\varepsilon_0}\int_V\rho dV\).

Author: Christian Cunningham

Created: 2024-05-30 Thu 21:17