Electrostatics Question 817
Question: For the configuration of media permittivities $ {\varepsilon _{0}},\varepsilon $ and $ {\varepsilon _{0}} $ between parallel plated each of area A, as shown in Fig. the equivalent capacitance
Options:
A) $ {\varepsilon _{0}}A/d $
B) $ \varepsilon {\varepsilon _{0}}A/d $
C) $ \frac{\varepsilon {\varepsilon _{0}}A}{d( \varepsilon +{\varepsilon _{0}} )} $
D) $ \frac{\varepsilon {\varepsilon _{0}}A}{( 2\varepsilon +{\varepsilon _{0}} )d} $
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Answer:
Correct Answer: D
Solution:
[d] $ C _{eq}=\frac{{\varepsilon _{0}}}{\frac{d}{K _{1}}+\frac{d}{K _{2}}+\frac{d}{K _{3}}} $ Here, $ K _{1}=K _{3}=1,K _{2}=\varepsilon /{\varepsilon _{0}} $
