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 difference, the distance between the capacitor plates is increase by `1.6 mm`. The dielectric constant of the plate is .

Options

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

Answer

A

Explanation

In air the potential difference between the plates
`V_(air) = (sigma)/(epsilon_(0)) .d` ..(i)
In the presence of partially filled medium potential difference between the plates
`V_(m) = (sigma)/(epsilon_(0)) (d-t+(t)/(K))` ..(ii)
Potential difference between the plates with dielectric medium and increased distance is
`V'_(m) = (sigma)/(epsilon_(0)) {(d + d') -t + (t)/(K)}` ..(iii)
According to question `V_(air) = V'_(m)` which gives
`K = (t)/(t -d')`
Hence, `K = (2)/(2 - 1.6) = 5`