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I read in my textbook that when two isolated capacitors are connected to each other, redistribution of charge takes place in such a way that they both acquire a common potential. My question is why charge redistribution should even happen, aren't the opposite charges on the plates holding up each other?

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  • $\begingroup$ How are they being connected? $\endgroup$ – Aaron Stevens Jan 23 '19 at 18:58
  • $\begingroup$ @AaronStevens the capacitor plates are connected to each other via non resistive wires. $\endgroup$ – Abirbhav Jan 23 '19 at 19:01
  • $\begingroup$ @Abirbhav See my revised answer. $\endgroup$ – Bob D Jan 24 '19 at 14:24
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First, if a plate from one capacitor is electrically connected to the plate of another capacitor by a wire, those two plates and wire must have the same potential. This is an elementary result from electrostatics.

If the other two plates are also electrically connected to each other by a wire, those two plates and wire must have the same potential.

It follows that the potential difference (voltage) across the two capacitors connected in this way must be the same.

Now, if there are initially two isolated capacitors that do not have the same potential difference, and if they are then connected together as I describe above, then there must charge redistribution at the time the capacitors are connected together.

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The plates that are connected together by a non-resistive wire are at the same potential, even though, as you said, the charges themselves are “locked in” since, with the other plates not connected, the charge cannot redistribute. Only when the other plates are connected together will the voltages across both capacitors be the same. Consider the following example. (Refer to figures below).

Fig A shows two isolated capacitors. Each + or – charge is a micro coulomb. So the charges and voltages for each capacitor are as shown, and are based on

$$Q=CV$$

Fig B shows the two capacitors’ positive plates connected together. Since the plates are connected together, they must be at the same potential. But the negative plates are not. If the voltmeter (VM) shown is connected across the negative plates, it will read as shown, based on Kirchoff’s voltage law (KVL).

Fig C shows both the positive and negative plates connected together. This allows electrons to flow between the plates so that, after equilibrium, the charges will redistribute themselves so that the voltages across both capacitors will be the same. In order to determine what the charges and voltage is, we apply the law of conservation of charge.

$$Q_{C1}+Q_{C2}= 6 + 4=10μC$$

$$\frac {Q_{C1}}{4}=\frac {Q_{C2}}{2}$$

Solving the equations simultaneously gives us

$$V_{C1}=V_{C2}=1.67 v$$

$$Q_{C1}=6.67 μC$$

$$Q_{C2}=3.33 μC$$

Hope this helps.

enter image description here

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  • $\begingroup$ Can you please elaborate how are you solving the equation ? i am not getting how you got 𝑉𝐶1=𝑉𝐶2=1.67𝑣 from 𝑄/𝐶14=𝑄/𝐶22.Also can you please tell me what will happen if i would have connected positive side of one capacitor with negative side of another capacitor $\endgroup$ – user232991 Jul 20 '19 at 3:16

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