How can Gauss's Law be Correct if We Consider Charges Not on The Gaussian Surface?

by modulus
Tags: charges, correct, gauss, gaussian, surface
 P: 103 Gauss' Law only figures out the the total charge enclosed by the Gaussian surface and on it, I treid this in a few cases where charges were placed only inside the surface, and my answers were correct. But, when I tried imagining how to do this if there were charges present outside the Gaussian surface , I figured the flux of the electric field would change, because charges outside the surface would vary the electric field lines of charges present inside the surface. So we won't be getting the flux due to charges only in the surface, but outside it too. That would give you the total charge of all charges around the surface, not only tose enclosed by the Gaussian surface, right?
 PF Gold P: 7,120 You get a flux from charges outside the surface, but it will all cancel out if there are no charges inside the surface.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,722 Any field lines from an outside charge that enter the volume enclosed by the surface will also exit, so the net flux from that charge will be zero.
 P: 103 How can Gauss's Law be Correct if We Consider Charges Not on The Gaussian Surface? Sorry for replying so late. I had gone out of town. OK, I understand that the electric field lines entering the surface will exit it. But, it will exit from the other side, correct? Now this brings up two problems: Suppose the Gaussian surface is irregular. Then, it wouldn't be neccessary that the electric field lines will exit in the same manner that they entered, so as to cancel out. Therefore, the total flux should be affected by the external field lines Also, if there are charges present inside the Gaussian surface, they should not allow the electric field lines due to the external charges to exit normally. Suppose there is a positive charge inside a cylindrical (symmetrical) surface and there is a positive charge outside of it too. Well then, the net electrical flux due to that charge would cancel out only if it exits fom the surface opposite to the one from where it entered. But, the positive charge inside would deflect the field lines of the positive charge outside, so as to prevent it from exiting from the opposite surface. This means the net electric flux would *not* be zero.
 Sci Advisor P: 2,751 Yes I think you have some misconception of Gauss's Law modulus. Gauss's Law simply states that the total flux leaving a closed surface is equal to the charge enclosed. The total flux is the surface integral (over the entire surface) of, $$\vec{D} \cdot d\vec{s}$$, it doesn't have to be symmetrical or uniform. ($D = \epsilon E$ btw).