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

Options

A. `2.67 xx 10^(-2)J`
B. `2.67 xx 10^(-4)J`
C. `3.67 xx 10^(-2)J`
D. `3.67 xx 10^(-4)J`

Answer

A

Explanation

Here, `C_(1) = 4 mu F = 4 xx 10^(-6)F, V_(1) = 200 V`
initial electrostatic energy stored in `C_(1)` is
`U_(1) = (1)/(2)C_(1)V_(1)^(2) = (1)/(2) xx 4 xx 10^(-6) xx 200 xx 200 = 8 xx 10^(-2) J`
When `4 mu J` capacitor is connected to uncharged capacitor of `2mu F`, charge flows and both acquire a common potential which is given by
`V = ("Total charge")/("Total capacity") , V = (C_(1)V_(1))/(C_(1) + C_(2)), V = (4 xx 10^(-6) xx 200)/((4+2) xx 10^(-6)) = (800)/(6)V`
`:.` Final electrostatic energy of both capacitors
`U_(2) = (1)/(2) = (C_(1) + C_(2))V^(2) = (1)/(2) xx 6 xx 10^(-6) xx (800)/(6) xx (800)/(6)`
`5.33 xx 10^(-2) J`
`:.` Energy dissipated in the form of heat and electromagentic radiation is
`U_(1) - U_(2) = (8 xx 10^(-2) - 5.33 xx 10^(-2))J = 2.67 xx 10^(-2) J`