Gauss Law for finite linear conductor Why can't Gauss Law be applied to a finite linear Conductor, while as the same can be applied to an infinite Conductor.
 A: The reasoning behind applying Gauss Law in infinitely large distributions of charge is symmetry. This is because in infinite charge contributions (say linear, i.e. 1 dimensional ones) there will always exist two infinitesimally small pieces of the charged rod (if it is a rod), one on the left and one on the right of the point at which we wish to calculate the electric field, whose contributions to the parallel with the rod component of the electric field will cancel.
In the case in which, say, you have a finite linear charge distribution (i.e. a rod again), and you try to calculate the electric field on top of the one end of the rod, the epxression you will get will not be accurate due to lack of symmetry (the charge distribution at the left hand side of the point at which you wish to calculate the electric field will not be the same as the charge distribution at the right hand side of the point). I do not exaclty recall right now all the details on the Gauss Law derivation (such that I can tell you whether or not it assumes symmetry as a requirement), but one way in which your approach will suffer is the fact that there is no way to calculate the component of the electric field that is parallel to the charge distribution, by using the Law of Gauss. So, you miss half of the answer before even starting to solve the problem. On the other hand, if your distribution was infinite, you wouldn't have this problem, as, according to all the above, the infinitesimally small charge on the one side of the point at which you wish to calculate the electric field, cancels the contribution to the non-vertical component of the electric field from the infinitesimally small charge on the other side. And since your distribution is infinite, you can always find such pairs!
Having said that, we conclude that it is still possible to apply Gauss law for charge distributions that are finite. You can do that if you want to derive an expression for the electric field on the symmetry axis that goes through the centre of the charged distribution, or alternatively, you can use Gauss law as an approximation to derive an expression for the electric field on points that are located very near the charged distribution (such that the rod length can be approximated as much larger than the distance from the rod at which one calculates the electric field)...
I hope this helps. If anything is not clarified, please comment.
A: The key in Gauss' law is that the field must be uniform over the Gaussian surface, i.e. it must have the same magnitude and direction on the surface.  In the case of the finite line, the field does not have constant magnitude or direction over the line: thus the field over a Gaussian cylinder would not be constant.
It is easy to convince yourself of this last point: if your Gaussian cylinder is close to one edge, or extends past one edge of the finite line, then surely the field there will not be the same as near the "middle" of the line.  In fact, you can compute the field "in front" of the charge line, and it will not have the same direction as the field near the middle of the line.
Of course, if the line is very long and you are interested in the field near the middle of your charged line, then the field would be approximately but very nearly constant over a Gaussian cylinder of finite length, and you can use Gauss' law to obtain the field near the center.
If the line is infinitely long, there is no end so the will be exactly uniform over the line.  Gauss' law is then very useful.
