A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness `d_(1)` and dielectric constant `K_(1)` and the other has thickness `d_(2)` and dielectric constant `K_(2)` as shown in figure. This arrangement can be through as a dielectric slab of thickness `d (= d_(1) + d_(2))` and effective dielectric constant `K`. The `K` is.
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Options

A. `(k_(1) d_(1) + k_(2) d_(2))/(d_(1) + d_(2))`
B. `(k_(1) d_(1) + k_(2) d_(2))/(k_(1) + k_(2))`
C. `(k_(1) k_(2) (d_(1) + d_(2)))/((k_(1) d_(2) + k_(2) d_(1)))`
D. `(2 k_(1) k_(2))/(k_(1) + k_(2))`

Answer

C

Explanation

The capacities of two individual condensers are `C_(1) = (k_(1) in_(0) A)/(d_(1))` and `C_(2) = (k_(2) in_(0) A)/(d_(2))`
The arrangement is equivalent to two capacitors joined in series. Therefore, the combined capacity `(C_(s))` is given by
`(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) = (d_(1))/(k_(1) in_(0) A) + (d_(2))/(k_(2) in_(0) A) = (1)/(in_(0) A) [(d_(1))/(k_(1)) + (d_(2))/(k_(2))] = (d_(1) k_(2) + d_(2) k_(1))/(in_(0) A K_(1) K_(2))`
or `C_(s) = (in_(0) A k_(1) k_(2))/(d_(1) k_(2) + d_(2) k_(1)) = (k in_(0) A)/((d_(1) + d_(2))) :. K = (k_(1) k_(2) (d_(1) + d_(2)))/((d_(1) k_(2) + d_(2) k_(1)))`