Flux through a closed surface containing no charge in non uniform electric field how can be zero? Imagine a cube which is placed right side of a point charge and it's electric field will be decreasing from left to right than how the net flux will be zero?
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$\begingroup$ That's the point of Gauss law, to give very fast an answer that would be tedious to obtain by hand. Can you point out where you're stuck in either of the computations? $\endgroup$– MiyaseCommented Jun 26, 2022 at 7:13
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$\begingroup$ Welcome to PSE and Gauss Law. Don't hope that you will learn from answers herein. You must study Classical Electrodynamics (Electrostatics first) and dig deep into the relevant part of differential geometry. $\endgroup$– VoulkosCommented Jun 26, 2022 at 7:14
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2 Answers
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Below is a diagram of a charge $q$ outside of a cube $ABCDEFGH$.
Certainly the magnitudes of the electric field entering face $ABCD$ of the cube are larger than those leaving the other five faces but to compensate for that the surface area of face $ABCD$ is less than that of the sum of the other five faces.
When the sums are done it is found that the electric flux entering the cube is equal to the electric flux leaving the cube.
In terms of electric field lines, any electric field line entering the cube must also leave the cube as there are no electric charges within the cube.
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Hope this video would help to understand: https://www.youtube.com/watch?v=jCuWaC0Oibw We may use the same trick for the cube.
