Condenser `A` has a capacity of `15 muF` when it is filled with a medium of dielectric constant `15`. Another condenser `B` has a capacity `1muF` with air between the plates. Both are charged separately by a battery of `100V`. After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is

Options

A. 400 V
B. 800 V
C. 1200 V
D. 1600 V

Answer

B

Explanation

For A,C `=15muF`
and `Q_(A)=CV=15xx10^(-6)xx100=15xx10^(-4)`C
For B, `Q_(B)=1xx10^(-6)xx100=10^(-4)C`
When the dielectric of A is removed, its capacity becomes
`(15xx10^(-6))/(15)=10^(-6)F`
Thus both will have the capacity of `1muF` each and since they are joined in parallel, `C_(P)=1+1=2muF`
Since the battery is removed, charges on A and B will remain the same.
`therefore Q_(A)=15xx10^(-4)C and Q_(B)=10^(-4)C`
`therefore`Common potential`=(Q_(A)+Q_(B))/(C_(1)+C_(2))`
`=(15xx10^(-4)+10^(4))/(2xx10^(-6))`
`therefore V=(16xx10^(-4))/(2xx10^(-6))=800V`