The battery has a potential difference. So, shouldn't battery generate an electric field directing from the positive side to the negative side? If that's the case, we know that potential in a wire(with wire having no resistance) stays constant unless a resistance comes across it. So, $\Delta V=0$. But we again know $\Delta V=Ed$ where $E$ is electric field and $d$ is the distance between the points whose potential difference we are measuring. So in a wire, since $\Delta V=0$, doesn't that mean $E=0$? So, how can electrons move through the wire if they feel no force i.e. $E$?
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2 Answers
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So, shouldn't battery generate an electric field directing from the positive side to the negative side?
That is indeed the case within the battery.
An electrochemical reaction within the battery moves charges within the battery so that one terminal has a deficiency of electrons (the positive terminal) and the other terminal has a surplus of electrons (the negative terminal). That migration of charges produced an electric field within the battery from the positive to the negative terminal. This electric field opposes the migration of the charges within the battery and, if the battery terminals are not connected externally, stops the migration so a steady potential difference is set up between the terminals of the battery.
When a conducting circuit is connected to the battery terminals then charges move externally and the electrochemical process within the battery tries to maintain the potential difference across the terminals of the battery by making charges migrate within the battery.
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The problem you are facing is based on an idealization of wires as perfect conductors. $\Delta V=0$ is true only if the resistivity of the wires is zero.
You have two ways to resolve this apparent paradox: either you consider a limit approach to infinite conductivity, where the electric field in the conductor $E=j/\sigma$ approaches zero in the limit, or you consider the small but finite conductivity of real wires.
In this latter case the field E in the wires is not zero. It's just very small and the voltage drop across any portion of wire is negligible.
In both cases, attaching a wire to a battery directly will result in exceedingly high current densities (infinite, in the limit for $\sigma$ that goes to infinite.)
I am not considering the actual case of a superconductor where the field inside would be zero and only a surface current will be allowed.
