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
Show Answer
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} $
