1.

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

Answer» Correct Answer - A
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`.


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