Capacitors

JEE Main level, numerical answer, medium

68 of 180

Question 68 of 180

A parallel plate capacitor is having separation between plates 0.8850.885 mm. It has a capacitance of 1 μ1\,\muF when the space between the plates is filled with an insulating material of resistivity 1×1013 Ω1\times 10^{13}\,\Omegam and resistance 17.7×1014 Ω17.7\times 10^{14}\,\Omega. Relative permittivity of the insulating material is α×107\alpha\times 10^7. The value of α\alpha is ________. (Take permittivity of free space =8.85×10−12=8.85\times 10^{-12} F/m)

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Answer: 2

The resistance of the insulating material is given by R=ρdAR = \dfrac{\rho d}{A}.

The capacitance of the parallel plate capacitor is given by C=ϵ0ϵrAdC = \dfrac{\epsilon_0 \epsilon_r A}{d}.

Multiplying the expressions for RR and CC, we get:

RC=ρϵ0ϵrRC = \rho \epsilon_0 \epsilon_r

Rearranging for relative permittivity ϵr\epsilon_r:

ϵr=RCρϵ0\epsilon_r = \dfrac{RC}{\rho \epsilon_0}

Substituting the given values:

ϵr=17.7×1014×10−61013×8.85×10−12\epsilon_r = \dfrac{17.7 \times 10^{14} \times 10^{-6}}{10^{13} \times 8.85 \times 10^{-12}}

ϵr=17.7×1088.85×101\epsilon_r = \dfrac{17.7 \times 10^8}{8.85 \times 10^1}

ϵr=2×107\epsilon_r = 2 \times 10^7

Comparing this with the given expression ϵr=α×107\epsilon_r = \alpha \times 10^7, we get:

α=2\alpha = 2

Answer: 22

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