Verified answer

When a voltage is applied to a capacitor, a charge is created on the plates (say Q).

The ratio of the charge (Q) to the voltage (say V) will give the capacitance of the capacitor.

C = Q/V or

Q = C*V

2 µF capacitor(say C1) is charged to a potential difference of 200v and then isolated:

hence the capacitor is charged (Q)

Q = 2*(10)^-6 * 200 = 400*(10)^-6

it is connected in parallel with a second capacitor :

C = C1 + C2 = 2.(10)^-2 + C2

The common potential becomes 40v (when the capacitors are connected in parallel):

Q = CV or (in our case C = C1+C2)

C = Q/V

C1+C2 = 400*(10)^-6 / 40

2*(10)^-6 + C2 = 400*(10)^-6/40

400*(10^-6 = 40C2 + 80*(10^-6

400*(10^-6 - 80*(10^-6 = 40C2

40C2 = 320*(10^-6 or

C2 = 320*(10^-6)/40 = 8*(10)^-6

Ans: C2 = 8 microFarad

The loss in charge in capacitor 1 is equal to gain in capacitor 2

The capacitors have same potential.

=>2x(200-40) = Cx40

now,solving for C, we get it as 8 microfarad.