Does there exist any electric field inside a charged conductor?

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Anindya Mondal
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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?
 
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Anindya Mondal said:
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
 
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Is it electric field or electric current?
 
Anindya Mondal said:
Is it electric field or electric current?
Both. The electric field drives the electric current.
 
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
 
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gau55 said:
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
 
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##.
 
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vanhees71 said:
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.
 
What don't you understand? It's very simple. In electrostatics by definition the current density vanishes. In a conductor, according to Ohm's Law, the current density is proportional to the electric field and thus the electric field must vanish within the conductor. I don't know, how I can this elaborate more.