SATHEE: Electrostatics Question 597

Electrostatics Question 597

Question: A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by [IIT-JEE Screening 2000]

Options:

A) $\frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{2 k_{3}}$

B) $\frac{1}{k} = \frac{1}{k_{1} + k_{2}} + \frac{1}{2 k_{3}}$

C) $k = \frac{k_{1} k_{2}}{k_{1} + k_{2}} + 2 k_{3}$

D) $k = k_{1} + k_{2} + 2 k_{3}$

Show Answer

Answer:

Correct Answer: B

Solution:

$C_{1} = \frac{K_{1} ε_{0} \frac{A}{2}}{(\frac{d}{2})} = \frac{K_{1} ε_{0} A}{d}$

$C_{2} = \frac{K_{2} ε_{0} \frac{A}{2}}{(\frac{d}{2})} = \frac{K_{2} ε_{0} A}{d}$ and $C_{3} = \frac{K_{3} ε_{0} A}{(\frac{d}{2})} = \frac{2 K_{3} ε_{0} A}{d}$

$\frac{1}{C_{e q}} = \frac{1}{C_{1} + C_{2}} + \frac{1}{C_{3}}$

$= \frac{1}{\frac{ε_{0} A}{d} (K_{1} + K_{2})} + \frac{1}{\frac{ε_{0}}{d} \times 2 K_{3}}$

$\frac{1}{C_{e q}} = \frac{d}{ε_{0} A} [\frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}]$

$C_{e q} = {[\frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}]}^{- 1} . \frac{ε_{0} A}{d}$ So $K_{e q} = {[\frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}]}^{- 1}$

Electrostatics Question 597

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Electrostatics Question 597

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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