The plates of a parallel plate capacitor are charged up to `100 v`. Now, after removing the battery, a `2 mm` thick plate is inserted between the plates Then, to maintain the same potential deffernce, the distance betweem the capacitor plates is increase by `1.6 mm`. The dielectric canstant of the plate is .

Options

A. `5`
B. `1.25`
C. `4`
D. `2.5`

Answer

A

Explanation

As battery is disconnected, so charge will remain the same. It is given that final potential is the same. So final capacitance should be equal to initial capacitance, i.e.
`(epsilon_(0)A)/d=(epsilon_(0)A)/((1.6+d)-t(1-1//K))`
or `K = 5`.