Three capacitors of capacitances 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 a 100 V supply.

Ans (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 C_eq the algebraic sum,

Therefore C_eq = C₁+C2 + C3 = 2 + 3+ 4 = 9pF

Therefore, total capacitance of the combination is 9 pF.

Ans (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… (i)

For C = 2 pF, charge = VC = 100 x 2 = 200 pC  = 2   x lO⁻¹⁰ C

For C = 3 pF, charge = VC = 100 x 3 = 300 pC   = 3   x lO⁻¹⁰ C

For C = 4 pF, charge = VC = 100 x 4 = 400 pC  = 4   x 10⁻¹⁰ C