Here, C1=C2=C3=9pF=9×10−12F
V=120volt
(a) Total capacitance of the series combination in given by
1C=1C1+1C2+1C3=3×19×10−12
=13×10−12
∴C=3×10−12F=3pF
(b) Let q be the charge on each capacitor. Then, sum of the potential differences across the plates of the three capacitors must be equal to 120 V i.e.,
V1+V2+V2=120
or qC1+qC2+qC3=120
or q9×10−12+q9×10−12+q9×10−12=120
or 3q9×10−12=120
or q=360×10−12C
Therefore, potential difference across a capacitor
=qcapacitance=360×10−129×10−12S=40V
Alternative method. Since the three capacitors are of the same capacitance, potential difference across each capacitor
=1203=40V.