electrostaticsgauss-lawhomework-and-exercises
I'm aware that location of charge enclosed within gaussian surface doesn't affects the net flux through the closed surface . But I wanted to know wether location affects flux through particular face.
Consider the case of a cube if the charge is place at centre then flux through each face will be equal . But what happens when it is not at center?
Does the face nearer to charge has greater flux?
In your last question it is important as to what you mean by
WRT the question.
If you are trying to find the E-field due to a point charge using Gauss then to make the surface integration easier you choose a surface which has the following properties:
the E-field direction is everywhere perpendicular to the surface
the E-field has a constant magnitude all over the surface.
In fact this is an equipotential surface.
Now if you move the Charge or the Gaussian surface you can still use Gauss but the surface integral becomes more difficult.
As the diagram shows the two conditions for a simple integration are not met in that the E-field is not perpendicular to the surface and it is not constant in magnitude.
In terms of flux it is perhaps easier to understand by going back in time when Faraday and others thought that the E-field lines (lines of force) actually "existed" and flux was a measure of the total number of these lines going through a surface. If you use that representation you can see that the number of field lines passing through the surface is the same in both cases, so the flux is the same.
So going back to your last question:
Also, would it matter (WRT the question) if the surface were a charged conductor, or simply imaginary?
In the first diagram if you introduced a spherical conductor where the Gaussian surface is shown (remember that it is an equipotential surface) the E-field would not change except become zero inside the conductor.
However arbitrarily adding a spherical conducting shell will change the E-field
Charge $-q$ is induced on the inside of the shell with a non-uniform distribution but the amazing thing (to me) is that the charge $+q$ which is induced on the outer surface is uniformly distributed on the surface of the sphere. Perhaps not so amazing if one thinks that those charges are trying to get as far away from one another as possible.
So perhaps it is safer to keep the Gaussian surface as a figment of your imagination?
Yes, for any closed Gaussian surface (including irregular surfaces), this law holds.
But we consider only symmetrical surfaces (sphere, cylinder, etc.) in academics as it is easy for performing the integration.
Let us consider the example which you have given:
Here, in order to find the net flux through the surface you must integrate each E.dA, which is not so easy. The advantage with symmetry is that E is always constant, and dA is always perpendicular to the surface so that you may integrate easily. Even in this case, if you integrate E.dA, you will end up with the same result.
Gauss law even applies for irregular surfaces :
The only problem is to integrate. You would have the same result.
How to think without mathematics :
As we know, flux is the measure of number of lines of electric field passing through the Gaussian surface. Even if you move the charge anywhere within the surface, the number of field lines won't change, right?
All the Gauss laws(of electrostatics, magnetism, gravitation), field vectors, inverse square laws can be derived from the GAUSS LAW OF DIVERGENCE. It's so exciting!!!
Best Answer
In the limit of a charge that is very close to one of the faces of your cube, it's obvious to see there are far more lines going through one face than the other:
You can conclude that the charge distribution, in general, will affect the flux through a particular surface. Do you see it now?