Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

Solution:

(a) Capacitances of the given capacitors are

$C_{1}=2 \mathrm{pF}$

$C_{2}=3 \mathrm{pF}$

$C_{3}=4 \mathrm{pF}$

For the parallel combination of the capacitors, equivalent capacitor $C^{\prime}$ is given by the algebraic sum,

$C^{\prime}=2+3+4=9 \mathrm{pF}$

Therefore, total capacitance of the combination is 9 pF.

(b) Supply voltage, V = 100 V

The voltage through all the three capacitors is same = V = 100 V

Charge on a capacitor of capacitance C and potential difference V is given by the relation,

$q=V C \ldots$ (i)

For $C=2 \mathrm{pF}$,

Charge $=V C=100 \times 2=200 \mathrm{pC}=2 \times 10^{-10} \mathrm{C}$

For $C=3 \mathrm{pF}$,

Charge $=V C=100 \times 3=300 \mathrm{pC}=3 \times 10^{-10} \mathrm{C}$

For $C=4 \mathrm{pF}$,

Charge $=V C=100 \times 4=200 \mathrm{pC}=4 \times 10^{-10} \mathrm{C}$