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Answers
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