Three capacitors each of capacitance $9 \;pF$ are connected in series.
$(a)$ What is the total capacitance of the combination?
$(b)$ What is the potential difference across each capacitor if the combination is connected to a $120 \;V$ supply?

(N/A) Capacitance of each of the three capacitors,$C = 9 \;pF$.
The equivalent capacitance $(C_{eq})$ of the combination of capacitors in series is given by the relation:
$\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} = \frac{3}{9} = \frac{1}{3} \;pF^{-1}$.
$\Rightarrow C_{eq} = 3 \;pF$.
Therefore,the total capacitance of the combination is $3 \;pF$.
$(b)$ Supply voltage,$V = 120 \;V$.
Since the capacitors are identical and connected in series,the potential difference $(V')$ across each capacitor is equal to one-third of the supply voltage:
$V' = \frac{V}{3} = \frac{120}{3} = 40 \;V$.
Therefore,the potential difference across each capacitor is $40 \;V$.