Three capacitors each of capacitance 9 pF are connected in series.

(a) Capacitance of each of the three capacitors, C = 9 pF

Equivalent capacitance (C’) of the combination of the capacitors is given by the relation,

$\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$

$\Rightarrow \frac{1}{C^{\prime}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{1}{3}$

$\Rightarrow C^{\prime}=3 \mathrm{pF}$

Therefore, total capacitance of the combination is 3 pF. 3 pF">3 p

(b) Supply voltage, V = 120 V

Potential difference (V') across each capacitor is equal to one-third of the supply voltage.

$\therefore V^{\prime}=\frac{V}{3}=\frac{120}{3}=40 \mathrm{~V}$

Therefore, the potential difference across each capacitor is 40 V.