# Charge outside Gaussian Surface doesn't contribute to Flux?

I roughly understand the explanation for this: any electric field line that enters the surface, must leave it, since field lines can't terminate abruptly in space. My question is, what if you have a charge $+2q$ outside your gaussian surface, and a charge $-q$ inside it? Wouldn't the field line end at the charge $+q$, and so the field lines that enter don't necessarily leave, making $+2q$ contribute a negative flux?

Wouldn't the field line end at the charge +q, and so the field lines that enter don't necessarily leave, making +2q contribute a negative flux?

The quantity of field lines that terminate on the charge -q is unchanged by the presence of charge elsewhere.

Imagine -q field lines terminate on the -q charge. Then, the net electric flux outward through the surface enclosing (only) the -q charge is $-\frac{q}{\epsilon_0}$.

If there is no other charge, the field lines of the charge extend to infinity.

If there is a 2q charge outside the surface, q field lines from the 2q charge terminate on the -q charge but there are still only -q field lines terminating on the -q charge.

In other words, whether the -q field lines originate at infinity or a 2q charge outside of the surface is irrelevant.

Consider the following diagram:

For either isolated charge, there are 16 field lines that originate or terminate on the charge.

If the two charges were not isolated from each other, the 16 field lines from the positive charge would terminate on the negative charge but there would still be only 16 field lines originating / terminating on either charge.

Thus, in either case, 16 field lines cross a surface enclosing just one of the charges - the electric flux through the surface is the same in either case.

The net flux crossing the Gaussian surface (according to Gauss's law) is irrespective of those charges outside it.

• The total flux leaving the surface drawn is -q (so it is actually entering the surface not leaving it). So although the 2q charge outside the surface do contribute to the flux crossing the surface, the net flux, according to his law, is irrespective of it. So what ever charge you put outside, the net flux will be determined by only those surrounded by the surface.

I think the correct answer for your question is that the electric charge outside of a closed Gaussian surface does not enter inside because the surface is enclosed. Take a look at the Faraday's cage and you can also take into account why we are not affected when driving in a car and get struck by a lightening.