Question

a parallel plate 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 intially in the single capacitior

Answer 1

The ratio should be 1:2.

When the capacitor is charged its energy is = 1/2×CV^2

=q^2/2C

Now,The charge should get equally distributed as the capacitance is the same.

So the energy becomes=(q/2)^2/2C

For two capacitor it is =2×(q/2)^2/2C

The initial energy/The combined energy

=q^2/2C/2×(q/2)^2/2C

=1/2

=1:2

Hope it helps. please mark it as brainliest

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 !!