Find the capacitance between A and B if two dielectric alabs (each of area `A`) of dielectric constants `K_(1)` and `K_(2)` and thicknesses `d_(1)` and `d_(2)` are inserted between the plates of a parallel plate capacitor of plate area `A`.
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Options

A. `C=Aepsilon_(0)[(K_(1)K_(2))/(K_(2)d_(1)+K_(1)d_(1))]`
B. `C=2Aepsilon_(0)[(K_(1)K_(2))/(K_(2)d_(1)+K_(1)d_(2))]`
C. `C=Aepsilon_(0)[(K_(1)K_(2))/(2K_(2)d_(1)+K_(1)d_(2))]`
D. `C=Aepsilon_(0)[(K_(1)K_(2))/(K_(2)d_(1)+K_(1)d_(2))]`

Answer

D

Explanation

The combination is equivalent to two capacitors connected in series
`1/(C)=1/(C_(1))+1/(C_(2))`
`=d_(1)/(AK_(1)epsilon_(0))+d_(2)/(AK_(2)epsilon_(0))=1/(Aepsilon_(0))[(K_(2)d_(1)+K_(1)d_(2))/(K_(1)K_(2))]`
or `C=Aepsilon_(0)[(K_(1)K_(2))/(K_(2)d_(1)+K_(1)d_(2))]`
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