The plates of parallel plate capacitor are charged upto $100\;V$. 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
Medium
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$(a)$ 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
${V_m} = \frac{\sigma }{{{\varepsilon _0}}}(d - t + \frac{t}{K})$ ..... $(ii)$
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 $K = \frac{2}{{2 - 1.6}} = 5$
${V_{air}} = \frac{\sigma }{{{\varepsilon _0}}}.d$ ..... $(i)$
In the presence of partially filled medium potential difference between the plates
${V_m} = \frac{\sigma }{{{\varepsilon _0}}}(d - t + \frac{t}{K})$ ..... $(ii)$
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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