Gauss' Law — Jemoka Knowledge Base

The Gauss’ Law is a principle of electric flux of uniformly distributed electric field along a surface: that, the electric flux through a closed surface is the sum of the electric charge enclosed divided by the permittivity of free space. That is:

\begin{equation} \oint E \cdot dA = \frac{\sum Q}{\epsilon_{0}} \end{equation}

somewhat motivating Gauss’ Law Consider a sphere with uniformly distributed charge on its surface. It has surface area 4 \pi r^{2}. Given the expression of electric flux and the fact that the origin change is in the center, and the test change is evenly distributed (i.e. E is held constant):

\begin{align} \Phi_{E} &= \int E \cdot dA \\ &= E\int dA \end{align}

Now, we are integrating across the entire surface of the sphere, so it is a closed integral. So:

\begin{align} \Phi_{E} &= \oint E dA \end{align}

We have the entire sum of the surfaces to be the surface area; so \oint dA = 4\pi r^{2}. Furthermore, recall that if the field is uniform, E is constantly at \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r^{2}}. So, substituting the two in:

\begin{align} \Phi_{E} &= \frac{1}{4\pi \epsilon_{0}} \frac{Q}{r^{2}} 4\pi r^{2} \\ &= \frac{Q}{\epsilon_{0}} \end{align}

where, \epsilon_{0} is the permittivity of free space. Congrats, we have Gauss’ Law: “the electric flux through the surface of an object is the sum of the charges enclosed divided by the permittivity of free space.” spheres electric field inside a closed conductor is zero This is a direct result of gauss’ law