Flux through an external charge

Problem

A point charge $q = +3\;\mu\text{C}$ is located outside a closed surface $\Sigma$ of arbitrary shape. Calculate the total flux of the electric field through $\Sigma$. Justify the result using Gauss’s theorem and explain why it does not depend on the shape of the surface.

Solution

Gauss’s theorem. The flux of the electric field through a closed surface depends only on the charge enclosed within it: $\Phi_\Sigma(\vec{E}) = \frac{Q_{\text{int}}}{\varepsilon_0}$

Enclosed charge. The charge $q$ lies outside $\Sigma$, so it does not contribute to $Q_{\text{int}}$: $Q_{\text{int}} = 0 \;\Rightarrow\; \Phi_\Sigma = \frac{0}{\varepsilon_0}$

Result. $\ev{\Phi_\Sigma(\vec{E}) = 0}$

Why it doesn’t depend on the shape. The field lines generated by the external charge cross the closed surface: every line that enters $\Sigma$ on one side leaves it on the other. The incoming flux (negative) and the outgoing flux (positive) exactly cancel, giving a total of zero. This balance holds regardless of the shape of $\Sigma$, because Gauss’s theorem counts only the enclosed charge and not the geometry of the surface: with no enclosed charges, the flux is always zero.

Links

Topics: Campo elettrico e potenziale Concepts: Teorema di Gauss Skills: Applicazione del teorema di Gauss Methods: Teorema di Gauss