# Does there exist any electric field inside a charged conductor?

Anindya Mondal
We know that there exists no electric field inside a conductor. But while calculating drift velocity of electrons inside an electric conductor, why do we consider the electrons are present inside the charged conductor?

Mentor
We know that there exists no electric field inside a conductor.
There certainly can exist an electric field inside a conductor. The electric field is proportional to the current density for ordinary conductors. This is known as Ohm's law

• FactChecker
Anindya Mondal
Is it electric field or electric current?

Homework Helper
Is it electric field or electric current?
Both. The electric field drives the electric current.

gau55
What you're referring to is probably what you get told in electrostatics at first, but the lack of an electric field is actually the condition for the static state, it can exist and as mentioned here causes a current to flow, this is now electrodynamics

• Anindya Mondal
Gold Member
On the atomic scale there are always significant electric fields but these average out.

Anindya Mondal
What you're referring to is probably what you get told in electrostatics at first, but the lack of an electric field is actually the condition for the static state, it can exist and as mentioned here causes a current to flow, this is now electrodynamics
Yeah, I refer to electrostatics

Gold Member
2022 Award
In electrostatics by definition you assume that all fields are time independent and that all current densities are vanishing, ##\vec{j}=0##. Now you have (in non-relativistic approximation) ##\vec{j}=\sigma \vec{E}##, where ##\sigma## is the electric conductivity of your medium. For a conductor ##\sigma \neq 0##, which implies that ##\vec{E}=0##, because in the electrostatic case you have by definition ##\vec{j}=0##.

• cnh1995
Anindya Mondal
In electrostatics by definition you assume that all fields are time independent and that all current densities are vanishing, ##\vec{j}=0##. Now you have (in non-relativistic approximation) ##\vec{j}=\sigma \vec{E}##, where ##\sigma## is the electric conductivity of your medium. For a conductor ##\sigma \neq 0##, which implies that ##\vec{E}=0##, because in the electrostatic case you have by definition ##\vec{j}=0##.
I can't understand, please be elaborate.