Electrostatics Question 639
A capacitor is filled with dielectrics. The dielectric resultant capacitance is
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:
[d] $ 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} $
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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