In a parallel plate capacitor the separation between plates is 3 x. This separation is filled by two layers of dielectrics, in which on layer has thickness x and dielectric constant ` 3k`, the other layer is of thickness `2x` and dielectric constant `5 k . It the plates of the capacitor are connected to a battery , then the ratio of potential difference across the dielectric layers is

Options

A. `1/2`
B. `4/3`
C. `3/5`
D. `5/6`

Answer

D

Explanation

Key Idea Capacitance of a parallel plate capacitor is give by relation.
`C= epsilon_(r).(epsilon_(0)A)/d`
where, `epsilon_(r)=` dielectric constant, A = area of parallel plate and d = distance between the plate.
Here, `d_(0) = 3x, d_(1) = x, epsilon_(1) = 3 k,d_(2) = 2 x and epsilon_(2) = 5k`
The capacitor is shown in the figure below,

Now , `C_(1) = 3 k(epsilon_(0)A)/x and C_(2) = 5k (epsilon_(0)A)/(2x)`
Since, m in series combination of capacitor, stored charge in each capacitor is same.
So, `V_(1) = Q/C_(1) and V_(2) = Q/C_(2)`
`rArr V_(1) = (thetax)/(3kepsilon_(0)A) and V_(2) = (theta 2 x)/(5 kepsilon_(0) A)`
The ratio, `V_(1)/V_(2) = ((thetax)/(3kepsilon_(0)A)) /( (theta 2 x)/(5 kepsilon_(0) A))= 5/(3xx 2) = 5/6`
Hence, option (d) is correct.