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}{2k _{3}} $
B) $ \frac{1}{k}=\frac{1}{k _{1}+k _{2}}+\frac{1}{2k _{3}} $
C) $ k=\frac{k _{1}k _{2}}{k _{1}+k _{2}}+2k _{3} $
D) $ k=k _{1}+k _{2}+2k _{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ C _{1}=\frac{K _{1}{\varepsilon _{0}}\frac{A}{2}}{( \frac{d}{2} )}=\frac{K _{1}{\varepsilon _{0}}A}{d} $
$ C _{2}=\frac{K _{2}{\varepsilon _{0}}\frac{A}{2}}{( \frac{d}{2} )}=\frac{K _{2}{\varepsilon _{0}}A}{d} $ and $ C _{3}=\frac{K _{3}{\varepsilon _{0}}A}{( \frac{d}{2} )}=\frac{2K _{3}{\varepsilon _{0}}A}{d} $
$ \frac{1}{C _{eq}}=\frac{1}{C _{1}+C _{2}}+\frac{1}{C _{3}} $
$ =\frac{1}{\frac{{\varepsilon _{0}}A}{d}(K _{1}+K _{2})}+\frac{1}{\frac{{\varepsilon _{0}}}{d}\times 2K _{3}} $
$ \frac{1}{C _{eq}}=\frac{d}{{\varepsilon _{0}}A}[ \frac{1}{K _{1}+K _{2}}+\frac{1}{2K _{3}} ] $
$ C _{eq}={{[ \frac{1}{K _{1}+K _{2}}+\frac{1}{2K _{3}} ]}^{-1}}.\frac{{\varepsilon _{0}}A}{d} $ So $ K _{eq}={{[ \frac{1}{K _{1}+K _{2}}+\frac{1}{2K _{3}} ]}^{-1}} $
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