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Well i think electric field must exist outside a current carrying conducting wire. Because if you consider a simple circuit with a resistor and a battery potential difference is created across the ends of the resistor as electric field exists inside the resistor. But what if i change my path? As electric field is a conservative field, Potential difference along any two points is independent of the path we take. I have heard from many sources and many books that current carrying wire have no field outside them as they are electrically neutral. But if there is no field and we take a path along air to reach the other end of the resistor, then potential difference must be 0 as we are walking through a field free space. So if electric field exists, charge must exist on the surface of conductor (charge can't stay inside the conductor). So does charge exist on the surface of conductor?

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TL;DR: Yes.

Any real resistive wire will have a potential difference along its path if current flows through it. This potential difference attracts charges to its surface.

Overall charge of your circuit with regards to the outer world determines whether the charges are positive on one end, or negative on the other end (or any combination of this).

Note that for usual circuits, these resistance-dependent charges are very small, as the voltage across a wire is typically in the mV range or less. In highly resitive wires like tunsten filament, or in high-frequency [EDIT] ... transmission lines where inductive impedance plays a major role, the electric field around the conductor can be substantial.

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Does charge develop on the surface of a current carrying wire?

Charge often develops on the surface of a current carrying wire. It may do so in the case of an external field, but it will commonly do so whenever the wire bends.

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    $\begingroup$ @Buraian thanks for the catch. :-) $\endgroup$ Jun 3, 2021 at 13:55

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