Electrostatics Question 561
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 be assumed to be in series combination.
BETA
