How can net field inside a conductor be zero at every point? What if a charged particle is at that point? The net electric field at any point inside a conductor which is not connected to an EMF source is zero and an external field is present. What is net electric field? It is the net field due to all the charges in the universe. Now consider an electron (or for that matter a proton in the nucleus of atoms) in a conductor. The net field on it is due to all the charges in the universe and hence due to it as well, but its field on itself is undefined ($r=0$ in the formula of field) so the net field inside a conductor where an electron lies must be undefined. But every book says that it is zero and to justify it they give the explanation that the conductor gets polarised and a field opposite to that of the external field gets generated and its magnitude is so that it exactly cancels it. But they don't consider the case of a charged particle being at the point so that it makes the net field on the point undefined.
 A: The electric field inside a conductor is zero.
Let's consider the alternative.  If there were a field at some point in the conductor that means there would be a force on the electrons there - and they would move.  They would redistribute themselves until the field became zero.
The books are talking about a scale bigger than the individual electrons, you are right that near each electron there would be a field directed towards each one.
It's similar to saying the sea or ocean is flat (when viewed from far away), if there were a hill of water in the ocean the weight of it would spread it out and it would become flat.
However oceans have waves and don't appear flat on a smaller scale.
A: Physically, they will continue to move, until there is no field.
Mathematically,
For a conductor following ohms law:
$J=\sigma E$
Taking the divergence of  both sides
$\nabla \cdot \vec{J} = \sigma \nabla \cdot E$
Sub in gauss law
$\nabla \cdot \vec{J} = \sigma \frac{\rho}{\epsilon_{0}}$
Envoking charge conservation
$\nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t}$
Substitue in
$-\frac{\partial \rho}{\partial t} = \sigma \frac{\rho}{\epsilon_{0}}$
$\int_{\rho_{0}}^{\rho} \frac{1}{\rho}\partial \rho = \int_{0}^{t} -\frac{\sigma}{\epsilon_{0}}\partial t$
$Ln(|\rho|) = -\frac{\sigma}{\epsilon_{0}} t + ln(|\rho_{0}|)$
$\rho = \rho_{0} e^{-\frac{\sigma}{\epsilon_{0}} t}$
Here  we can see that in an UNBOUNDED conductor, following ohms law, the charge density  everywhere vanishes, leaving zero electric field. This is extremely similar to "large" bounded conductors where charge accumulates on the boundary
