Three capacitors of capacitance `2 pF, 3 pF and 4 pF` are connected in parallel. (a) what is the total capacitance of the combination ? (b) Determine the charge on each capacitor, If the combination is connected to 100V supply.

Explanation

(a) capactiances of the given capacitors are
`C_(1)=2pF`
`C_(2)=3pF`
`C_(3)=4pF`
For the parallel combination of the capacitors, equivalent capacitors C' is given by the algebric sum,
`C'=2+3+4=9 pF`
Therefore, total capacitance of the combination is 9 pf
(b) supply voltage , V=100 V
The voltage through all the three capacitor is same V=100 V
Charge on the capacitor of capacitance C and potential difference V is given by the relation,
q=Vc...(i)
For C=2pF
Charge =VC =`100xx2=200 pC=2xx10^(-10)C`
for C=3pF,
Charge =VC =100 x 3 =300 pC =`3xx10^(-10)C`
For C=4pF, ltBrgt charge =VC=100 x 4 =200 pC `=4xx10^(-10) c`