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2. Electric Potential and Capacitance
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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
A$5$
B$1.25$
C$4$
D$2.5$
Solution
$(a)$ 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 $K = \frac{2}{{2 – 1.6}} = 5$
Standard 12
Physics
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