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While applying Gauss' Law, the electric field at a point on the Gaussian surface has to come from superposition of electric fields of all the charges, whether outside or inside the Gaussian surface. However, the charge $Q$ has to be only from inside the surface. The law somehow implies that the flux due to a charge is only dependent on the charge inside ($\frac{Q_{in}}{\epsilon_0}$) but the other side of equation has a term ($E$) which depends on outside charges as well.

This is a very common cliché for books and sources to mention as a fact that the electric field is due to all charges but $Q$ is only charge inside. But none of them, so far, as I have read, seem to tell why. Why is it so? Isn't it trivial because in a sense we're getting the same flux whether the external charge exists or not?

Rew
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  • does this answer your question: https://physics.stackexchange.com/questions/126294/why-can-charges-outside-be-ignored-in-gausss-law – Brain Stroke Patient Jul 11 '20 at 06:40
  • @BrainStrokePatient Yes it partially does but my question is that is not trivial to use $E$ field from the outside charges when their flux is zero? Couldn't I just do without them? I'm hoping for a counterexample where ignoring the Electric field due to outside charges DOES affect the answer. – Rew Jul 11 '20 at 06:53
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    It is trivial. You can always ignore the electric field due to outside charges. In fact, in Gauss's law problems, you would normally solve for the electric field anyway. And what you get is the electric field due to the enclosed charges, not the net electric field on the surface. – Brain Stroke Patient Jul 11 '20 at 06:57
  • @BrainStrokePatient But from what the books mention, it is implied that the electric field should be the net electric field due to all charges, isn't it? – Rew Jul 11 '20 at 07:13
  • No I don't think any of the books I read implied that. Even if some book did, it doesn't matter as the flux due to outside electric fields is zero anyways. So taking the net electric field or taking the electric field only due to the enclosed charges will yield the same flux. – Brain Stroke Patient Jul 11 '20 at 07:19
  • @BrainStrokePatient indeed. Thanks mate – Rew Jul 11 '20 at 07:24

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