Question

solve the above question ​

Answer 1

Answer:

electrostatic energy lost in the process is 6 x 10^-6

Explanation:

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

E = 1/2CV^2

= 1/2 * (600 * 10^-12) * ( 200 )^2

=1.2 * 10^-5 J

If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C') of the combination is given by,

1/C' = 1/C + 1/C = 1/600 + 1/600 = 2/600 = 1/300

C' = 300pF

New electrostatic energy can be calculated as

E' = 1/2 * C' * V^2 = 1/2*300*(200)^2 = 0.6 * 10^-5

Loss in electrostatic enegy = E - E' = 1.2 x 106-5 - 0.6 x 10^-5 = 0.6 x 10^-5 = 6 x 10^-6 J