When a voltage source is applied across an ideal wire, do the push of electric field of battery source causes the accumulation of charges on the surface of wire or the capacitance of wire causes it?
One more doubt regarding this is - if we have a straight ideal (no turns at all) wire, somehow if I generate electric field parallel to the wire through a battery source (so the circuit is closed) will there be accumulation of charges( along the wire length) to cancel electric field of battery in this case? If so how?
Answer
Ideal wire cannot have net electric field inside. Whatever electric field the source can create, is cancelled out in the ideal wire by charges around, mostly on the terminals and wire surface. These negative charges are always present there, but usually their field is cancelled out by equal amount of charge of positive sign. When you connect the voltage source, the negative charges move a little so their field is no longer completely cancelled out by the positive charges. This motion is due to electromotive force of the battery and mutual push of the negative charges on the surfaces.
Re the second question, yes, the fact that the wire is ideal means the charges on the wire will redistribute to cancel out the external field of the battery. This is only an approximation to what would happen in reality. In reality, the wire is not ideal and the voltage source is not ideal either. There would be large current, there would be non-zero electric field in the wire and the battery would heat up and its voltage would drop down. It would still be above zero, as long as the current is flowing. Which in case of shorted battery won't be long - don't try to short out any battery, you would destroy it and it may result in dangerous explosion.
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