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?

Explanation

Capacitance of each of the three capacitor, C = 9pF Equivalent capacitance `C_(s)` of the series combination of the capacitors is,
`(1)/(C_(s))=(1)/(9) +(1)/(9)+(1)/(9) =(3)/(9)`
`:. (1)/(C_(s))=(1)/(3)`
`:. C_(s)= 3pF`
(b)
Capacitance of each of the three capacitors is same and all capacitors are in series. Suppose, the potential difference across `C_(1) , C_(2)` and `C_(3)` are `V_(1 ), V_(2)` and `V_(3)` respectively
`:. V= V_(1)+V_(2)+V_(3)`
`:. 120 = 3V_(1)` or `3V_(2)` or `3V_(3)[ because V_(1) = V_(2)=V_(3)]`
`:. V_(1) = 40 V, V_(2)=40 ` V and `V_(3) = 40 `V
Second method for (b) :
For series combination
`V= V_(1)+V_(2)+V_(3)`
Suppose the charge on each capacitor is q
`120 =(q)/(C_(1))+(q)/(C_(2))+(q)/(C_(3))`
Now `C_(1),C_(2),C_(3)` all are same
`:. 120 = (3q)/(C)= (3q)/(9xx10^(-12))`
`:. C_(1) = C_(2)=C_(3)= C= 9xx10^(-12)F `
`:. q = 360 xx10^(-12)` C
The potential difference across each capacitor
`V_(1)=V_(2) =V_(3)=(q)/(C)`
`= (360xx10^(-12))/(9xx10^(-12))`
40 V