A `4muF` capacitor is charged by a 200 V supply. It is then disconnected from the supply . And is connected to another uncharged `2muF` capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?

Options

A. `3.43 xx 10^(-2)`
B. `2.67 xx 10^(-2)`
C. `2.67 xx 10^(-4)`
D. `3.43 xx 10^(-4)`

Answer

(b)

Explanation

Energy charge stored at capacitor ,`q=C_(1)V_(1)` = `4 xx 200 = 800 mu `C
Common potential on both capacitors is ,
V =`C_(1)V_(1) + C_(2)V_(2)/(c_(1) + C_(2)) = 800/6 V`
Loss in energy = Initial energy - Final Energy
`= 1/2 C_(1)V_(1)^(2)- 1/2 (C_(1) + C_(2)V^(2)`
=`1/2 xx 4xx 10^(-6) xx (200)^(2) - 1/2 (4 + 2) xx 10^(-6) (800/6)^(2)`
` = 8 xx 10^(-2) - 64/12 xx 10^(2) = 2.67 xx 10^(-2)`