Question

A capacitor of 200 pF is charged by a 300 V battery. The battery is then disconnected and the charge capacitor is connected to another uncharged capacitor of 100 pF. Calculate the difference between the final energy stored in the combined system and the initial energy stored in the single capacitor.

Answer 1

Given C = 200 pF = 200 × 10⁻¹² F

V = 300 V

The energy (initial) stored by the capacitor is

`U_i = 1/2 CV^2`

     `= 1/2 xx 200 xx 10^-12 xx 300 xx 300 = 9 xx 10^-6 j`

The charge on the capacitor when charged through 300 V battery is

Q = CV

= 200 × 10⁻¹² × 300

= 6 × 10⁻⁸ C

When two capacitors are connected, they have their positive plates at the same potential and negative plates also at the same potential. Let V’ be the common potential difference. By charge conservation, charge would distribute but total charge would remain constant;

Thus ,

`Q =q +q'`

`q/C = q/C'`

`q/200 = q/100`

`q =2q'`

Thus,

`Q = 2q'+q' =3q'`

So,`q' =Q/3 =(60nC)/3 = 20nC`

  `q=2q' =40nC`

Thus, `U_f=q^2/(2C) +(q^'2)/(2C')`

             `=1/2 xx (40 xx 10^-9)^2/(200 xx 10^-12) + 1/2xx ((20 xx 10^-9)^2)/(100 xx 10^-12)`

             `=(4 xx 10^-6 + 2 xx 10^-6)`

            ` = 6 xx 10^ -6 j`

Difference in energy = final-initial

`=U_f -U_i`

`= 6 xx 10^-6 -9 xx 10^-6`

`=-3 xx 10^-6`

Thus, difference in energy is 3 × 10⁻⁶ J.