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.
.

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

The capacitance of parallel plate capacitor filled with dielectriic block has thickness `d_(1)` and dielectric constant `K_(2)` is given by
`C_(1)=(K_(1)epsi_(0)A)/(d_(1))`
Similarly, capacitance of parallel plate capacitor filled with dielectric lbock has thickness `d_(2)` and dielectric constant `K_(2)` is given by
`C_(2)=(K_(2)epsi_(0)A)/(d_(2))`
Since, the two capacitors are in series combination, the equivalent capacitance is given by
`(1)/(C)=(1)/(C_(1))+(1)/(C_(2))`
or
`C=(C_(1)C_(2))/(C_(1)+C_(2))=((K_(1)epsi_(0)A)/(d_(1))(K_(2)epsi_(0)A)/(d_(2)))/((K_(1)epsi_(0)A)/(d_(1))+(K_(2)epsi_(0)A)/(d_(2)))=(K_(1)K_(2)epsi_(0)A)/(K_(1)d_(2)+K_(2)d_(1))` . . .(i)
But the equivalent capacitances is given by
`C=(Kepsi_(0)A)/(d_(1)+d_(2))`
On comparing we have
`K=(K_(1)K_(2)(d_(1)+d_(2)))/(K_(1)d_(2)+K_(2)d_(1))`.