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Electric potential is related to Electric Field by negative integral of Field with distance.If charges move away from the positive terminal (Assuming conventional flow) why is there no drop in electric potential even though the charges are moving in the direction of electric field?

Some of the answers I read for the similar question said that the electrons in wires rearrange themselves in a way to make the electric field uniform in the wire.However,this doesn’t explain the absence of potential drop in the wires (due to electric field) since the field is still there.

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Real wires do indeed have an electric field through them, and as such, there is a voltage drop associated with a current of electrons. This is totally fine, because real wires have nonzero resistance. As such, a voltage drop is expected.

The "wires" used in circuit diagrams are assumed to have zero resistance, and zero voltage drop across them. They are abstractions that do not correspond to physical objects, but are rather used to show solely how the other components of the circuit are connected.

The reason this is acceptable is because wires typically have very low resistance, so low that the voltage drop across them under normal circumstances is negligible. As such, we can treat them as zero-resistance objects for the purposes of circuit analysis. It's a simplifying assumption that will not give you the correct physics whenever the resistance of the wires matters. Examining the electric field and voltage drop inside wires is one of the situations in which the resistance of the wire matters, so this makes sense.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Oct 19 '18 at 16:08

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