# Why do charges outside the closed Gaussian surface not matter for the flux in the event they end inside the surface?

I think I am misunderstanding something elementary. Supposing a scenario where you have a negative charge enclosed in a Gaussian surface, why would the presence of outside charges not affect the flux?

In scenario 1 there are no outside charges, so there will be field lines coming in from infinity and none going out so total flux is negative, which makes sense. But in scenario 2 where you have multiple positive charges outside the Gaussian, do field lines from these charges not end in the negative charge in the Gaussian? And the more positive charges you have, the more field lines that end in the negative charge? I don't understand why the flux is the same in both scenarios.

And the more positive charges you have, the more field lines that end in the negative charge?

The number of field lines chosen to describe the amount of charge is purely arbitrary and doesn't follow any physical laws. They are simply used to depict the magnitude of charge on individual charges.

But in scenario 2 where you have multiple positive charges outside the Gaussian, do field lines from these charges not end in the negative charge in the Gaussian?

Field lines are nothing but the superposition of existing vector force fields and can easily be resolved to their original form. In other words, the Electric field of individual charges is independent of each other and so is the flux. Flux due to outside charges cancels out since the flux entering = flux leaving and we are left with flux of enclosed charges.

Electric field obeys a superposition principle. If you have the situation of an isolated negative charge inside a closed surface, then indeed field lines come into the surface and end on the negative charge.

If you now introduce other charges outside the surface, they of course affect the total electric field, but the new field can be decomposed into that due to the original negative charge plus the field due to the new charges.

The electric field lines due solely to the new charges will pass in(out) of the closed surface and then out(in) again. As a result, none of those field lines will end inside the surface and thus the additional net flux in or out of the closed surface, due to the new charges, is exactly zero.

Actually, this is a very profound question. Depending on perspective you might redefine the outside of the surface as the new "inside" and the inside as the new "outside". With this, you might think the Gauss law should still apply, and thus solely depend on the charges that were originally outside.

To put things in perspective, consider a fixed negative charge enclosed, as in your example. Outside, around the surface, we put a positively charged shell with the same net charge. Now, all field lines have a beginning and an end, and cross the surface. We know by the Gauss law that the total flux across the surface does not depend on the precise location of the inner charge within the surface. Likewise, we can expand the outer shell around the surface, which will displace the field lines a bit but the flux accross the surface remains fixed.

This means that we can move the outside charge arbitrarily far away from the surface, and again it doesn't have a net influence on the flux. Finally, consider the limit where you bring the outside positive shell to the boundary of the universe. Here you got the result.