# Electric field near a conducting surface

while calculating electric field near the surface of a conductor, we take only a small patch of charges near the point and draw a Gaussian surface in the form of a cylinder to calculate the field, I do not understand why we ignore the other charges on the surface of the conductor.

• It's a good question but could have been formulated better. Mar 9, 2017 at 15:25

Consider the electric field in a small region near the surface of a conductor in electrostatic equilibrium. The electric field in this region has only a normal component to the surface. Otherwise the tangential component would accelerate the free charges in disagreement of the initial hypothesis of equilibrium.

Since we are interested in the field near the surface we use a small Gaussian surface. if we choose a small cylinder, it is easy to compute the flux. There is only flux across the top of the cylinder. After getting the result we are only able to infer the value of the electric field in that small region. If we were allowed to extend the Gaussian to arbitrary large regions, the result for the electric field would be extended as well. But in the example of an arbitrary shaped conductor, we cannot consider a large Gaussian surface. The field would cross the Gaussian surface in complicated ways that we would not be able to compute the flux.

• what i dnt understand is not the functioning of gauss's law here,why we take that much of charge only where as there are many charges through out the conductor,its an arbitrarily shaped conductor and is absolutely unsymmetrical,the books just take a small portion of the conductor and are using gauss's law,are we neglecting the other charges,thanks for ur support. Mar 9, 2017 at 14:55
• @sachinrath123 I am not sure what your doubt is. In any case, I rewrote my answer. Mar 9, 2017 at 17:31
• I'm also not sure what the problem is. Mar 10, 2017 at 0:05

Well, you would draw a cylindrical Gaussian surface for a straight conducting line (or cylinder) of linear density $\lambda = Q/L$.
If it is an infinite line, then you take a small patch because of the symmetry of the problem.
[Of course, it if weren't infinite you would have boundary problems at the ends of the conductor but that is Classical Electrodynamics problem.]
Furthermore you know that the field inside the conductor is 0 at equilibrium and that the electric field is always normal to the surface because if it weren't, charges would be compelled to move and you would no longer have an electrostatics problem.
Summa summarum, you need not take the whole conductor because of the symmetry and the fact that only on the outside you will have an electric field $\neq0$

EDIT 1:
You can neglect the other charges because the Gauss law will be the same for them. If you have a mutated potato conductor, it doesn't matter what the charge distribution is. You know that in an electrostatic problem, charge will distribute itself on the surface in such a way that in equilibrium there is no movement (hence statics). Therefore, no matter which portion you chose, Gauss's law will be the same.

Gauss law is the simplest way of handling when it comes to charges and field flux. As I have said, if you have a potato conductor, it makes no sense to choose the whole Gaussian surface as one small patch is enough to calculate the field because you know that the charges redistribute along the surface so that $E_{inside}≠0$ . It would be impossible to integrate the whole potato.

• what i dnt understand is not the functioning of gauss's law here,why we take that much of charge only where as there are many charges through out the conductor,its an arbitrarily shaped conductor and is absolutely unsymmetrical,the books just take a small portion of the conductor and are using gauss's law,are we neglecting the other charges,thanks for ur support. Mar 9, 2017 at 14:55
• Well you should have said that then. Give info when you ask a question pretty please. You can neglect the other charges because the Gauss law will be the same for them. If you have a mutated potato conductor, it doesn't matter what the charge distribution is. You know that in an electrostatic problem, charge will distribute itself on the surface in such a way that in equilibrium there is no movement (hence statics). Therefore, no matter which portion you chose, Gauss's law will be the same. Mar 9, 2017 at 15:23
• m sorry for less information,i have been working on the same since two months and m coming to a conclusion that gauss law is just a formality in physics as without that all situations can be handled,if we bother a very normal situation that there are so many charges around a place and we are taking only a small patch of it thinking that there are no charges nearby,it almost makes no sense as if we use couloumbs law that is just integration of E.A ,all the charges are to be taken into account,can we ignore gauss law from here and work on it in a more simple way,thanks. Mar 9, 2017 at 16:58
• Gauss law is the simplest way of handling when it comes to charges and field flux. As I have said, if you have a potato conductor, it makes no sense to choose the whole Gaussian surface as one small patch is enough to calculate the field because you know that the charges redistribute along the surface so that $E_{inside} \neq0$. It would be impossible to integrate the whole potato. Mar 10, 2017 at 0:02
• thanks a lot Dominik Car for ur explanation,the confusion is cleared,it was actually the title of the section in the book what says "electric field near the surface of a conductor" and that is what puzzled me,instead of mentioning "electric field near the surface of a conductor",it should be "electric field near the surface of a conductor by the small part of the conductor near the point not of the whole conductor",thanks. Mar 10, 2017 at 20:23