Why isn't the electric field infinite at a conductor surface?

[parsa7parsa]

Hi
We know that the electric field at the surface of a conductor only have a normal component equal to ρ /ε (finite number).
But let’s consider the point P (at the surface of a conductor ) . Assume that there is a charge at an infinitesimal distance from the point p . we can obtain the field at the P by the fourmula (E=Kq/r) .obviously, E ~1/r. so the normal component of the field is infinite. Now if we add the field due to other charges, it will remain infinite. So where could I be possibly wrong?

 

[Simon Bridge]

We know that the electric field at the surface of a conductor only have a normal component equal to ρ /ε (finite number).

It's the gradient of the electric field that has that value.

The charges in the conductor will respond to the electric field of the small charge close to the conductor - affecting the way the total field comes out. How do they respond?

Note - at a very small distance from a point charge, the field is not infinite.
If the charge is actually at point P, then it is part of the conductor. Inside a conductor, the charges are infinitesimally small (in this model).

 

[ManDay]

we can obtain the field at the P by the fourmula (E=Kq/r)

You should ask yourself what exactly is "q" in that equation going to be for your capacitor with a given charge density (*cough*) ρ

 

[parsa7parsa]

please note that the charge is <within> the conductor