The capacitance of a variable capacitor joined with a battery of 100 V is changed from 2`mu`F to `10 mu F ` . What is the change in the energy stored in it ?

Options

A. `2xx10^(2) J`
B. `2.5 xx10^(2)` J
C. `6.5 xx10^(-2)J` J
D. `4 xx10^(-2)` J

Answer

A, B, D

Explanation

The energy stored in capacitor of 2`mu`F.
`U_(1)= (1)/(2) xx2xx10^(-6)xx(100)^(2) =10^(-2)` J
The energy stored in capacitor of `10 mu F `
`U_(2) = (1)/(2)xx10xx10^(-6)xx(100)^(2)`
`U_(2) = 5xx10^(-2)` J
`:.` The change in stored energy
`U = U_(2)-U_(1)= 5xx10^(-2) - 1 xx10^(-2)`
`:. U = 4xx10^(-2) `J