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I understand that if there are different capacitors of different capacitance then they can store same charge at different voltage differences only. But I wanted to understand that why the above happens when the capacitors are arranged in series...I want to know the mechanism behind that if we arrange different capacitors in series then they store same charge at different potential differences.....why they store same charge only that is flowing in the whole circuit due to net potential difference of the circuit....why they don't store different charges whose sum is equal to the net charge flowing in the circuit?

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Suppose an uncharged capacitor with plates $a$ and $b$ and another uncharged capacitor with plates $c$ and $d$.

Plate $b$ is connected to plate $c$ to form a series arrangement of the two capacitors with plates $a$ and $d$ to be connected to an external circuit.

The total charge on plates $b$ and $c$ is zero and there is no conducting path to the "outside world", thus the total charge on the two plates $b$ and $c$ must always be zero.

Now connect the series capacitor to an external voltage source so that plate $a$ now has a charge of $+q$ on it.

The charge $+q$ on plate $a$ induces a charge of $-q$ on plate $b$.

Since the total charge on the two plates $b$ and $c$ is zero an charge of $+q$ must be induced on plate $c$ which in turn induces a charge of $-q$ on plate $d$.

Thus each capacitor stores the same amount of charge all because the total charge on plates $b$ and $c$ must always be zero.

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  • $\begingroup$ @Farcher....the charge coming on the plates should be at the inner sides so that the formed electric field will be bounded....right? $\endgroup$ Commented Aug 12, 2023 at 14:15
  • $\begingroup$ If I am right then +Q charge on the farther most plate which is connected directly to positive terminal of the battery should have +Q/2 on either sides of the plate so that electric field inside the field should remain zero so how can the whole +Q charge come to the inner side of the plate and not on both the sides as +Q/2.... kindly explain me thank you! $\endgroup$ Commented Aug 12, 2023 at 16:59
  • $\begingroup$ The assumption I made, but did not state, was that the battery was connected to both sides if the capacitor. $\endgroup$
    – Farcher
    Commented Aug 12, 2023 at 19:01
  • $\begingroup$ can you pls explain the charge distribution by making a circuit with two capacitors in series and both the ends are connected to the battery. Because I wrote my previous comments assuming that you have connected the circuit of two capacitors in series with a battery. $\endgroup$ Commented Aug 21, 2023 at 5:57
  • $\begingroup$ @SukritiSharma As requested, I have added a diagram to my answer. $\endgroup$
    – Farcher
    Commented Aug 21, 2023 at 7:04
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Since a circuit has a potential difference, there is a surplus of electrons on one side and a shortage of electrons on the other. A capacitor in this circuit means the interruption of this circuit. However, electrons from the surplus side are pushed once onto the connected capacitor plate and (almost) as many electrons from this side are pushed into the potential sink on the other capacitor side.

With two capacitors in series, more electrons can never be displaced in this way than the small capacitor allows.

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