# Equivalent capacitance in series

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)?

• The equivalent capacitance in this case would be $Q/2$, not $Q$, I believe, if each capacitor has charge $Q$. – Zack Hutchens May 28 '18 at 20:34
• "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 $W = \frac{1}{2}\frac{Q^2}{C}$ – Alfred Centauri May 28 '18 at 21:15

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!