Net Electric field inside a conductor=0

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varadgautam
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If the net electric field inside a conductor is zero, how come electrons flow on applying potential difference across it? What force acts on the electrons? We've been taught that
an electric field E sets up, so the force on electrons is eE (e=charge on electron).
acceleration a=eE/m (m=mass of electron)
and then they calculated drift velocity. But by Gauss' Law, Electric field inside a conductor is zero, so thre must be no force on the electrons.
 
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a) that is only for a perfect conductor :)

b) given a perfect conductor, the electrons will travel only on it's surface, not inside it.
 
[tex]\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}[/tex]
(where [itex]\rho[/itex] is the charge density)
so this means that the divergence of the electric field is constant within the conductor. Surely this allows a constant electric field throughout the conductor?

Edit: I'm being stupid. This would only allow a constant electric field if the total charge density is zero. But I guess this is true in conductors where there are both positive and negative charges.
 
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In electrostatics, i.e., in the case of time-independent fields and charge distribution and the absense of currents, the electric field in conductors, no matter whether they are ideal or resitive, must vanish because of Ohm's Law,

[tex]\vec{j}=\sigma \vec{E}.[/tex]

Since [itex]\vec{j}=0[/itex] for [itex]\sigma \neq 0[/itex] one must have [itex]\vec{E}=0.[/itex]

In any other cases, one has to determine the electric field inside conductors by the general boundary and initial conditions, necessary for the unique solution of Maxwell's Equations.
 
So to directly answer the OP: If a conductor has no currents flowing through it, then there is no electric field inside the conductor. And if there is a current flowing, there generally will be an electric field.
This sound right to everyone?