SATHEE: Electrostatics Question 64

Electrostatics Question 64

Question: The capacitance of an air capacitor is $15 μ F$ the separation between the parallel plates is $6 m m$ . A copper plate of $3 m m$ thickness is introduced symmetrically between the plates. The capacitance now becomes [MP PMT 1995]

Options:

A) $5 μ F$

B) $7.5 μ F$

C) $22.5 μ F$

D) $30 μ F$

Show Answer

Answer:

Correct Answer: D

Solution:

By using $C_{a i r} = \frac{ε_{0} A}{d},$

$C_{m e d i u m} = \frac{ε_{0} A}{d - t + \frac{t}{K}}$

For K = ¥ $C_{m e d i u m} = \frac{ε_{0} A}{d - t}$

therefore $\frac{C_{m}}{C_{a}} = \frac{d}{d - t}$

therefore $\frac{C_{m}}{15} = \frac{6}{6 - 3}$

therefore $C_{m} = 30 μ C$

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Electrostatics Question 64

Question: The capacitance of an air capacitor is 15μF the separation between the parallel plates is6mm . A copper plate of 3mm thickness is introduced symmetrically between the plates. The capacitance now becomes [MP PMT 1995]

Options:

Answer:

Solution:

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Question: The capacitance of an air capacitor is $15 μ F$ the separation between the parallel plates is $6 m m$ . A copper plate of $3 m m$ thickness is introduced symmetrically between the plates. The capacitance now becomes [MP PMT 1995]