Question

The plates of a parallel plate capacitor of capacity 50 F are charged to a potential of 100 volts and disconnected then the plates are separated from each other so that the distance between them is doubled. How much is the energy spent in doing so
1) 2 25 10-2 J 2) 12.5 10-2 J
3) 2 25 10-2 J 4 ) 12.5 10 -2J

Answer 1

Explanation:

We know that,

W = 1/2 CV²

 Initial work = 1/2 x 50 x 10⁻⁶ x 10⁴

If  distance increase by 2 , C decrease by 2 and as battery is diconnected V increase by 2  

 Final work = 1/2 x 50 x 10⁻⁶ x 10⁴ x 2

            w =  final - initial

Net work = 1/2 x 50 x 10⁻⁶ x 10⁴

                 =  25x 10⁻²

Hope you understand

This is Habel Sabu ..... Isn't it the Brainliest