Electrostatics Question 602

Question: A capacitor is filled with dielectrics. The resultant capacitance is [UPSEAT 2001]

Options:

A) $ \frac{2{\varepsilon _{0}}A}{d}[ \frac{1}{k _{1}}+\frac{1}{k _{2}}+\frac{1}{k _{3}} ] $

B) $ \frac{{\varepsilon _{0}}A}{d}[ \frac{1}{k _{1}}+\frac{1}{k _{2}}+\frac{1}{k _{3}} ] $

C) $ \frac{2{\varepsilon _{0}}A}{d}[ k _{1}+k _{2}+k _{3} ] $

D) None of these

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Answer:

Correct Answer: D

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}{2d}=\frac{K _{3}{\varepsilon _{0}}A}{2d} $ Now, $ C _{eq}=C _{3}+\frac{C _{1}C _{2}}{C _{1}+C _{2}} $

$ =( \frac{K _{3}}{2}+\frac{K _{1}K _{2}}{K _{1}+K _{2}} ).\frac{{\varepsilon _{0}}A}{d} $



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