When is the Electric Flux zero through a closed surface? I know that if number of field lines entering is equal to leaving then it is true. Which I think is always true for a closed surface. So can there be any situation in which an isolated surface has 0 flux?

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    $\begingroup$ Gauss's law states that the flux through a closed surface is proportional to the charge contained within the surface... $\endgroup$ – lemon Apr 23 '16 at 13:25
  • $\begingroup$ So What ? Will that answer my question $\endgroup$ – Kumar Sambhav Apr 23 '16 at 13:37
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    $\begingroup$ Yes... To put it even more simply: the flux through a closed surface is equal to the charge contained within it (in the right units). So under what circumstance will the flux be zero? $\endgroup$ – lemon Apr 23 '16 at 14:02

As stated by Lemon, electric flux through a volume enclosed by a closed surface is zero when the volume contains no net charge.

Electric flux through a closed surface $\rm S$ is $$\Phi= \int_{\mathrm S} \,\mathbf E\cdot \mathbf n\,\mathrm d^2 \mathbf r\;.$$

Now, according to Divergence Theorem,

\begin{align}\int_{\mathrm S} \,\mathbf E\cdot \mathbf n\,\mathrm d^2\mathbf r &= \int_{\mathrm V}\, \mathbf \nabla \cdot \mathbf E \,\mathrm d^3\mathbf r\\ &= \int_{\mathrm V}\, \frac{\rho(\mathbf r)}{\epsilon_0}\,\mathrm d^3 \mathbf r\\ &= \frac{Q_\textrm{int}}{\epsilon_0}\tag 1\end{align}

From $(1)\,,$ it can be concluded that when $Q_\textrm{int}= 0\,$ the electric flux is zero.


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