A capacitor of capacitance `4muF` is charged to a potential of 100 V. it is then disconnected from the battery and connected in parallel with another capacitor `C_(2)`. If their common potential is 40 volts, then the value of `C_(2)` is

Options

A. `2muF`
B. `3muF`
C. `5muF`
D. `6muF`

Answer

D

Explanation

`Q_(1)=C_(1)V=4xx10^(6)xx100=400xx10^(-6)C and V'=40V`
`therefore C_(1)+C_(2)=(Q)/(V')=(400xx10^(-6))/(40)=10xx10^(-6)F`
`therefore C_(2)=10xx10^(-6)-4xx10^(-6)=6xx10^(-6)F`