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We all know from a course on electrostatics that the electric field inside a perfect conductor, placed inside a static electric field, is zero. Now imagine the same perfect conductor is connected to a DC voltage source by perfectly conducting wires. What is the electric field inside the conductor in such a situation?

Since $\vec{E}=\rho\vec{j}$, and resistivity $\rho=0$ for a perfect conductor, here too, the electric field inside seems to be zero. Also since $E=(V_A-V_B)/\ell$, and since there is no voltage drop between the points $A$ and $B$ in the absence of resistance, there is no electric field.

But isn't this odd? If the electric field inside is zero, what causes the electrons to move through the conductor and maintain a steady current?

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  • $\begingroup$ physics.stackexchange.com/questions/404609/… $\endgroup$
    – BowlOfRed
    Commented Feb 7 at 6:28
  • $\begingroup$ Despite what is usually taught in electrostatics as an illustration, the question as to what is inside a perfect conductor makes no sense because a perfect conductor is not a electromagnetic body but is a boundary condition to a set of partial differential equations. We illustrate its meaning by the verbiage around it regarding free charges and some such but it is a boundary condition not a body within that boundary. When that boundary is to enclose a physical body then in the simplest case we model it with no free charges using the equation $J=\rho E$ and assume that $\rho >0$. $\endgroup$
    – hyportnex
    Commented Feb 7 at 10:49

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"Now imagine the same perfect conductor is connected to a DC voltage source by "perfectly conducting wires." That is called a short circuit. There will initially be a huge electric field and a huge current, until the wire explodes, or the capacitor explodes, or a fire starts, or a fuse saves the situation. Don't try that at home.

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