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While solving some simple capacitor problems, I found that certain "excess" capacitors could be removed from a circuit if they're connected across equipotential wires (especially in case of symmetrical circuits), with the logic that no current flows through equipotential surfaces.

WHY?

One of my friends asked me to conduct a thought experiment.

Keep some water at same levels in 2 tubs. Now connect their bases. Does water flow in between them? Do levels change?

They don't, unless - and I might be wrong - we consider (random) Brownian motion of the water molecules. Water (it should, as far as my knowledge goes) flows randomly everywhere, and thus into the connector pipe that joins two tubs too, and hence in between the two tubs too.

If all I said was true, water maintains a dynamic equilibrium in its level across the two tubs. But nothing says it doesn't flow.

Same way for current. I understand that no work is done when objects are moved across equipotential surfaces. But does that mean, they WON'T go at all by themselves?

P.S. - Is this expulsion of any chance of random movements all because charges are classically considered to be not random?

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Commented May 31, 2020 at 21:27

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Note to myself:

When we are talking about current, as in current doesn't flow across equipotential surfaces, what we mean is net current.

No net current flows through equipotential surfaces. This is perfectly comfortable with the idea of a dynamic equilibrium and hence the concept of perpetual random motion. Current does flow, but whatever does is neutralised by an oppositely incoming current. So overall it is zero.

Thanks to @Krishna for the help.

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  • $\begingroup$ Are you talking about the case of ideal wires in circuits? If so, then saying there is no net current through the wires is incorrect. Or maybe I don't understand your question. $\endgroup$
    – BioPhysicist
    Commented May 31, 2020 at 17:43
  • $\begingroup$ I'm talking about ideal wires connected across 2 equipotential surfaces, yes. $\endgroup$
    – BeBlunt
    Commented May 31, 2020 at 17:58

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