a 2uF capacitor is charged to a potential of 200v and then isolated. When it is connected in parallel switch a second capacitor which was initially not charged, the common potential becomes 40v. The capacitance of the second capacitor is?

Answer 1

energy stored = 1/2 * capacitance * [(voltage)^2]

capacitors in parallel equal the sum of the individual capacitances

1/2 * 2 * 200^2 = 1/2 * (2 + c) * 40^2

solve for c