Question

Deduce an expression for the effective capacitance of capacitors of $C_1,C_2$ and $C_3$ connected in series

Answer 1

Capacitors in series: Consider three capacitors of capacitance $C_1,C_2$ and $C_3$ connected in series. Let $V$ be the potential difference applied across the series combination. Each capacitor carries the same amount of charge $q$. Let $V_1,V_2,V_3$ be the potential difference across the capacitors $C_1,C_2,C_3$ respectively. Thus $V=V_1+V_2+V_3$.
The potential difference across each capacitor is,
$V_1=\frac{q}{C_1};V_2=\frac{q}{C_2};V_3=\frac{q}{C_3}$;
$V=\frac{q}{C_1}+\frac{q}{C_2}+\frac{q}{C_3}=q[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}]$
If $C_S$ be the effective capacitance of the series combination, it should acquire a charge $q$ when a voltage $V$ is applied across it.
$V=\frac{q}{C_S}=\frac{q}{C_1}+\frac{q}{C_2}+\frac{q}{C_3}$
$\therefore \frac{1}{C_S}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$