The plates of parallel plate capacitor are charged up to 100V . A 2 mm thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6 \, mm$ . The dielectric constant of the plate is
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In air the potential difference between the plates
$V_{air} = \frac{\sigma}{\varepsilon_0} . d$ ..... (i)
In the presence of partially filled medium potential difference between the plates
$V_{m} = \frac{\sigma}{\varepsilon_{0}} \left( d - t + \frac{t}{K} \right)$ ..... (ii)
Potential difference between the plates with dielectric medium and increased distance is
$V_m' = \frac{\sigma}{\varepsilon_0} \left\{ (d + d') - t + \frac{t}{k} \right\}$ ..... (iii)
According to question $V_{air} = V_{m}'$ which gives $K = \frac{t}{t - d'}$
Hence $K = \frac{2}{2 - 1.6} = 5$
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