Question

A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Answer 1

Let 'q' be the charge on the charged capacitor. Energy stored in it is U=q22C When another similar uncharged capacitor is connected, the net capacitance of the system is C' = 2C
The charge on the system is constant. So, the energy stored in the system now is U'=q22C'⇒U'=q222C⇒U'=q24C Thus, the required ratio is U'U=q24Cq22C⇒U'U=12

Answer 2

Hey !!

Energy stored in a capacitor

                 U = 1/2 CV² (or) 1/2 q²/C

Capacitance of the parallel combination

              = C + C = 2C

Here, the total charge, Q remains the same

∴ Initial energy = 1/2 q²/C

and Final energy = 1/2 q²/2C

∴ Ratio of energies =  Final energy / Initial energy

           = 1/4.q²/C / 1/2

           = 2/4

            = 1/2

GOOD LUCK !!