Electrostatics Question 561

Question: Two dielectric slabs of constant $ K _{1} $ and $ K _{2} $ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor [MNR 1985; MP PET 1999; DCE 2002]

Options:

A) $ \frac{2{\varepsilon _{0}}A}{2}(K _{1}+K _{2}) $

B) $ \frac{2{\varepsilon _{0}}A}{2}( \frac{K _{1}+K _{2}}{K _{1}\times K _{2}} ) $

C) $ \frac{2{\varepsilon _{0}}A}{2}( \frac{K _{1}\times K _{2}}{K _{1}+K _{2}} ) $

D) $ \frac{2{\varepsilon _{0}}A}{d}( \frac{K _{1}\times K _{2}}{K _{1}+K _{2}} ) $

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Answer:

Correct Answer: D

Solution:

The two capacitors formed by the slabs may assumed to be in series combination.



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