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 $
BETA
