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As far as I know, Gaussian surface's are surfaces that enclose a net charge and are shaped in such a way that the electric field vector's magnitude is the same over the entirety of the surface. My textbook states that the introduction of an external charge Q will not affect the net flux of the surface but it may affect the distribution of the field lines. Will the inclusion of Q affect the magnitude of E at points on the surface due to superposition and, if so, won't that affect the electric field vector used in the integral to calculate the net flux of the surface? Also, if that's the case then will the Gaussian surface no longer be Gaussian or is the magnitude of E at the surface due exclusively to the enclosed net charge the only constant?

Thanks

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  • $\begingroup$ You've got something wrong: the electric field magnitude does not have to be constant on a Gaussian surface. Often the location and shape of a Gaussian surface is chosen to make it so, but that is only to ease the subsequent math. Any closed surface placed anywhere can be a Gaussian surface. $\endgroup$ – garyp Sep 15 '15 at 1:25
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When you have a charge next to a conducting surface, you induce movement of charge on the surface of the conductor. The net result has to be that a) the field lines are perpendicular to the conductor, and b) Gauss's law still holds.

For the simple example of a charge $q$ next to a plane surface, we can use the method of an image charge: the boundary condition (a) can be satisfied by imagining a second charge $-q$ on the other side of the plane, such that the boundary conditions at the conductor surface are satisfied. We know from the uniqueness theorem that is this A solution, it must be THE solution - so we can use this to estimate the new electric field distribution. While it is perpendicular to the surface, its magnitude changes with position. If you were to take a small Gaussian volume at one point on the surface, you would see a net electric field from which you conclude that locally there is charge buildup (image credit: wikipedia:

enter image description here

The integral over the whole conductor cannot change because the net charge is still zero - but locally the field can and does change, and so does the charge distribution.

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