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?

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

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

1/C_eq =1/C + 1/C + 1/C = 3/C = 3/9 = 1/3

=> 1/C_{eq =} 1/3 =>  C_{eq =} 3 pF

Therefore, total capacitance of the combination is 3 pF.

(b) Supply voltage, V = 100 V

Potential difference (V₁) across each capacitor is equal to one-third of the supply voltage.

Therefore V₁=V/3=120/3=40V

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