Question

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.

Answer 1

$\text{Part (a)}$

$\text{Step 1: Equivalent capacitance [Refer Fig.1]}$

As all capacitances are in parallel

Therefore, $C_{eq}=C_1+C_2+C_3$

$=(2+3+4)pF=9pF$

$\text{Part (b)}$

$\text{Step 2: Charge on Each Capacitor [Refer Fig. 2]}$

Since all the capacitance are in parallel, Therefore voltage across each will remain same $=100V$

So Charge on each capacitor is given by:

$Q_1=C_{1}V=2pF\times 100V=200pC$

$Q_2=C_{2}V=3pF\times 100V=300pC$

$Q_3=C_{3}V=4pF\times 100V=400pC$