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I'm struggling a bit understanding capacitors. I understand all the basics/formulas of capacitors in series and in parallel, but is lacking a complete intuitive understanding of how capacitors. So I came upon a question the other day with a charged capacitor (by a voltage source) and then removing the battery and adding an uncharged capacitor, the charge flows along as if this was in a parallel sense, thus making the situation like figure 1 below.

Correct me if I'm wrong, so I thought the charge flows along, creating the dipoles like shown (pos to pos/ neg to neg). What if we now had more capacitors connected in series? When no battery is connected, would it create the dipoles like in fig 2 or in fig 3? (I am getting awfully confused...) enter image description here

And also I thought that when batteries are involved in the circuit, perhaps the dipoles are created based on the direction of current? Is that what's happening?

Thanks in advance. It would also be helpful if you can provide an intuitive understanding of what capacitance/capacitor is.

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1 Answer 1

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If we start some uncharged capacitors in series in a circuit with no source of potential difference, they stay uncharged.

If we start some charged capacitors in series with an open circuit (for instance, by charging them with a battery, then opening the circuit and removing the battery), they'll look like the capacitors in figure 2 (but with the circuit open). If the only circuit element between two points is a wire, they are at the same potential.

enter image description here

In ideal circuit diagram land, if we close the circuit, we get a divide-by-zero error for current (finite potential difference divided by resistance 0), square it and multiply by zero for power dissipated across the wire (resistance 0), and get undefined. In reality, wires have finite resistance, so we have by Ohm's law:

$P = V^2/R$

a nontrivial number squared divided by a very small number is a very much bigger number than the thermal power a real wire can radiate at temperatures compatible with solidity, so a split second later our circuit looks like this:

enter image description here

With a little bit of residual + charge on the left and - charge on the right.

Since vaporized wires are an excellent insulator, we've opened the circuit again.

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  • $\begingroup$ How do you get the situation in your first figure? What is the source of this charge distribution? $\endgroup$
    – nasu
    Feb 16, 2023 at 11:42
  • $\begingroup$ @nasu I'd have to charge them in a differently designed circuit or circuits and connect them that way after they were charged. $\endgroup$
    – g s
    Feb 16, 2023 at 16:16
  • $\begingroup$ There's nowhere in the first figure that one could just replace a length of wire with a battery to charge them that way but I don't think there's anything unphysical about the distribution itself. On reflection, that means that my last sentence in my answer is wrong; I'll delete it. $\endgroup$
    – g s
    Feb 16, 2023 at 16:23

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