Let two capacitors be connected in series across a potential difference and each stores a charge Q. Then, while deriving equivalent capacitance, why are we considering that the equivalent capacitor will store a charge Q and not 2Q (which is the total charge)?

  • $\begingroup$ The equivalent capacitance in this case would be $Q/2$, not $Q$, I believe, if each capacitor has charge $Q$. $\endgroup$
    – zh1
    May 28, 2018 at 20:34
  • $\begingroup$ "each stores a charge Q" - capacitors don't store charge - both capacitors are electrically neutral - capacitors store energy. For a capacitor of capacitance $C$, the voltage across is given by $V = \frac{Q}{C}$ where it is understood that $Q$ is the charge on one plate while the other plate has charge $-Q$. The energy stored is $W = \frac{1}{2}\frac{Q^2}{C}$ $\endgroup$ May 28, 2018 at 21:15

1 Answer 1


In a sense a capacitor stores zero charge, even when it's charged, as the charges on its plates are equal and opposite! This might seem a silly pedantic point, but it's relevant to your question. Better to think of the charge that you use in the equation $Q=CV$ as the charge of one sign that flows on to one plate and off the other, that is the charge that seems to flow through the capacitor!

Now let's apply this to your two capacitors in series. Suppose one capacitor has plates A and B and the other has plates C and D, with B connected to C.

If plate A has a positive charge, Q, then the charges on the plates are

A: +Q, B: –Q, C: +Q, D: -Q.

The two plates that are connected together from an 'island'; they have no net charge.

If the capacitor combination were contained in a black box with just two terminals, one connected to A and the other to D, if charge Q seemed to flow through the box, that is if charge -Q went into one terminal and out of the other, there'd be charge ±Q on each capacitor!


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