Electrostatics Question 48

Question: The plates of parallel plate capacitor are charged upto$ 100\ V $ . A$ 2mm $ 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 [MP PMT 1991]

Options:

A) 5

B) 1.25

C) 4

D) 2.5

Show Answer

Answer:

Correct Answer: A

Solution:

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}}}{ (d+d’)-t+\frac{t}{K} } $ ….. (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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