UY1: Usage Of Gauss’s Law

Negative Charge Enclosed

If the surface enclose a negative charge, net electric flux through the surface enclosing the negative charge is negative.

$$\Phi_{E} = \frac{1}{\epsilon_{0}} q < 0$$

No Charge Enclosed

When the surface does not enclose any charge, total net electric flux is zero. This is due to the magnitude of electric flux entering the surface is the same as the magnitude of electric flux leaving the surface.

Multiple Charges Enclosed

Suppose a Gaussian surface encloses not just one point charge q but several charges $q_{1}$, $q_{2}$, $q_{3}$, …

The total electric flux through the Gaussian surface is equal to the total (net) electric charge enclosed by the surface, divided by $\epsilon_{0}$.

$$\Phi_{E} = \frac{Q_{encl}}{\epsilon_{0}}$$

$$Q_{encl} = q_{1} + q_{2} + q_{3} + …$$

Short proof:

\begin{aligned} \Phi_{E} &= \oint \vec{E}.d\vec{A} \\ &= \oint \left( \sum\limits_{i} \vec{E}_{i} \right) . d\vec{A} \\ &= \oint \vec{E}_{1} . d \vec{A} + \oint \vec{E}_{2} . d \vec{A} + \oint \vec{E}_{3} . d\vec{A} + … \\ &= \frac{Q_{encl}}{\epsilon_{0}} \end{aligned}

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