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 thought as a dielectric slab of thickness d = `( d_(1) + d_(2) )` and effective dielectric constant k. The k is ......


(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) `(2K_(1)K_(2))/(K_(1)+K_(2))`

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. `(2K_(1)K_(2))/(K_(1)+K_(2))`

Answer

C

Explanation

Capacitance of a parallel plate capacitor filled with dielectric of constant `K_(1)` and thickness `d_(1)` is
`C_(1) = (K_(1) in_(0)A)/(d_(1))`
and for dielectric `K_(2)` and thickness `d_(2)` the capacitance of a parallel plate capacitor
`C_(2) = (K_(2) in_(0)A)/(d_(2))`
Both capacitors are in series so equivalent capacitance,
`(1)/(C) = (1)/(C_(1))+(1)/(C_(2))`
`:.C= (C_(1)C_(2))/(C_(1)+C_(2))=(((K_(1)in_(0)A)/(d_(1)))((K_(2)in_(0)A)/(d_(2))))/((K_(1)in_(0)A)/(d_(1))+(K_(2)in_(0)A)/(d_(2)))`
`:. C = (K_(1)K_(2)in_(0)A)/(K_(1)d_(2)+K_(2)d_(1))`
So multiply the numberator and denominator by `(d_(1)+d_(2))`
`C=((K_(1)K_(2)in_(0)A))/((K_(1)d_(2)+K_(2)d_(1)))xx(in_(0)A)/((d_(1)+d_(2)))`
Effective capacitance ,
`C = (K in_(0)A)/((d_(1)+d_(2))`
Comparing equation (2) and (3)
`K = (K_(1)K_(2)(d_(1)+d_(2)))/((K_(1)d_(2)+K_(2)d_(1)))`