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 :

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_(1)+k_(2)d_(2)))`
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)epsilon_(0)A)/(d_(1))`
Similarly, for other capacitance of a parallel plate capacitor filled with dielectric of constant `k_(2)` and thickness `d_(2)` is,
`C_(2)=(k_(2)epsilon_(0)A)/(d_(2))`
Both capacitor are in series so equivalent capacitance C is related as :
`(1)/(C)=(1)/(C_(1))+(1)/(C_(2))=(d_(1))/(k_(1)epsilon_(0)A)+(d_(2))/(k_(2)epsilon_(0)A)`
`=(1)/(epsilon_(0)A)[(k_(2)d_(1)+k_(1)d_(2))/(k_(1)k_(2))]`
So, `C=(k_(1)k_(2)epsilon_(0)A)/((k_(1)d_(2)+k_(2)d_(1)))` ....(i)
`C.=(kepsilon_(0)A)/(d)=(kepsilon_(0)A)/((d_(1)+d_(2)))` .....(ii)
where, `d=(d_(1)+d_(2))`
Comparing eqns. (i) and (ii), the dielectric constant of new capacitor is :
`k=(k_(1)k_(2)(d_(1)+d_(2)))/((k_(1)d_(2)+k_(2)d_(1)))`