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enter image description here

In the above image, one capacitor is uncharged while the other is at a potential Vi. If we close the switch then potential difference across both capacitor would become equal(because the upper and lower plates of both capacitors respectively would gain same potential)

My question is, why does the switch need to be closed? Sure, the circuit would remain open. But if we consider, say, 2 separately charged conductors and connect them with a copper wire, charge flows till both are at the same potential. Why cannot this happen with the capacitors in the picture? The bottom plate of both the capacitors are still connected, irrespective of the switch being closed or opened. Still, charge doesn't flow between the plates. Why is this?

My thought process was that in the charged capacitor, initially the net potential on the positive plate is actually the potential difference between both plates. And this plate is connected with the bottom plate of the uncharged capacitor. So, since there is a difference between both plates(Vi and zero), charge should flow.

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  • $\begingroup$ Rh said he stopped replying to you bc youve never once upvoted or green checked anyones answer. $\endgroup$
    – Al Brown
    Aug 24 at 9:02
  • $\begingroup$ You are right and I apologize. But the website says I need at least 15 reputation to vote. $\endgroup$
    – Adi
    Aug 24 at 9:42
  • $\begingroup$ Also if I may ask, who is rh? $\endgroup$
    – Adi
    Aug 24 at 9:42
  • $\begingroup$ What do you mean by "the other capacitor is at potential Vi? The whole point of a capacitor is that it has two plates which are insulated from each other, and are (usually) at different potentials. $\endgroup$
    – alephzero
    Aug 24 at 15:53
  • $\begingroup$ By that I meant the potential difference between the plates of a capacitor $\endgroup$
    – Adi
    Aug 24 at 17:52
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There are two ways to look at this

One is that opposite charges attract each other. So if the bottom half (A) is charged relative to ground, and even more charged relative to the top left half (B), then whatever charges we find in A will be most attracted to B and will migrate as close as possible to them by electrostatic attraction.

We could imagine A is positively charged, B negatively, and the upper right, C, is uncharged. The charges in A are most attracted to B as C has no net charge, so they aren’t attracted to C.


The other way to look at this is that whenever charges are free to move, they “run downhill” to lower energy potential states. The charges will migrate to the configuration of lowest potential. The entire section A and entire section B will still be charged wherever the charge is, to they do comprise a capacitor. So where do the charges go to minimize $V$ across this A - B capacitor?

To minimize $V= \frac{q}{C}$ we maximize $C$, because $q$ is not changing. Because $C=\varepsilon_0 \frac{A}{d}$ for a plate capacitor, we can therefore deduce that maximizing $C$ means lowering the distance between charge (because for a plate maximizing $C$ would mean lowering $d$).

But we also know that potential is the work per unit charge to create this configuration. To minimize potential we minimize how much $\int F dx$ would be used to set-up the separated charges. Clearly they should be as close as possible for that. They’d all be on the left plates.

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  • $\begingroup$ Thanks. Actually, please correct me if my thinking is wrong. I assumed that some charge (say, half of the charge from the positive plate) has flown. Now the original capacitor has different charges on both the plates. Thus, now there is a net electric field towards positive plate which originally lost charge. This field will cause the charges to return (since current is produced due to electric field) to the plate until the field outside becomes zero, cause the charges to stop flowing. $\endgroup$
    – Adi
    Aug 24 at 9:40
  • $\begingroup$ I’ll check your question tmrw. Gnight $\endgroup$
    – Al Brown
    Aug 24 at 10:19

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