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

(a) Capacitances of the given capacitors: C₁ = 2 pF, C₂ = 3 pF and C₃ = 4 pF
For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum,
Therefore Ceq = C₁ + C₂ + C₃ = 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 capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation, q = VC
For C = 2 pF, charge = VC = 100 × 2 = 200 pC = 2 × 10^–10 C
For C = 3 pF, charge = VC = 100 × 3 = 300 pC = 3 × 10^–10 C
For C = 4 pF, charge = VC = 100 × 4 = 400 pC = 4 × 10^–10 C