SATHEE: Electrostatics Question 48

Electrostatics Question 48

Question: The plates of parallel plate capacitor are charged upto $100 V$ . A $2 m m$ thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6 m m$ . 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

$V_{a i r} = \frac{σ}{ε_{0}} . d$ ….. (i)

In the presence of partially filled medium potential difference between the plates $V_{m} = \frac{σ}{ε_{0}} (d - t + \frac{t}{K})$ ….. (ii) Potential difference between the plates with dielectric medium and increased distance is $V_{m}^{'} = \frac{σ}{ε_{0}} (d + d^{'}) - t + \frac{t}{K}$ ….. (iii)

According to question $V_{a i r} = V_{m}^{'}$ which gives $K = \frac{t}{t - d^{'}}$ Hence $K = \frac{2}{2 - 1.6} = 5$

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Electrostatics Question 48

Question: The plates of parallel plate capacitor are charged upto100 V . A2mm thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by1.6 mm . The dielectric constant of the plate is [MP PMT 1991]

Options:

Answer:

Solution:

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Question: The plates of parallel plate capacitor are charged upto $100 V$ . A $2 m m$ thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6 m m$ . The dielectric constant of the plate is [MP PMT 1991]