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

Question

In a parallel plate capacitor the separation between plates is 3 x. This separation is filled by two layers of dielectrics, in which one layer has thickness x and dielectric constant 3 k, the other layer is of thickness 2 x and dielectric constant 5 k. If the plates of the capacitor are connected to a battery, then the ratio of potential difference across the dielectric layers is (A) (1/2) (B) (4/3) (C) (3/5) (D) (5/6)

Tardigrade Admin · Accepted Answer

Capacitance of a parallel plate capacitor is give by relation,

C = ε_{r} \cdot \frac{ε _{0} A}{d}

where,

ε_{r} =

dielectric constant,

A =

area of parallel plate and

d =

distance between the plate.
Here,

d_{0} = 3 x, d_{1} = x, ε_{1} = 3 k

,

d_{2} = 2 x

and

ε_{2} = 5 k

The capacitor is shown in the figure below,

Now,

C_{1} = 3 k \frac{ε _{0} A}{x}

and

C_{2} = 5 k \frac{ε _{0} A}{2 x}

Since,

m

in series combination of capacitor, stored charge in each capacitor is same.
So,

V_{1} = \frac{Q}{C _{1}}

and

V_{2} = \frac{Q}{C _{2}}

\Rightarrow V_{1} = \frac{θ x}{3 k ε _{0} A}

and

V_{2} = \frac{θ 2 x}{5 k ε _{0} A}

The ratio,

\frac{V _{1}}{V _{2}} = \frac{\frac{θ x}{3 k ε _{0} A}}{\frac{θ 2 x}{5 k ε _{0} A}}

= \frac{5}{3 \times 2} = \frac{5}{6}

$\frac{1}{2}$

$\frac{4}{3}$

$\frac{3}{5}$

$\frac{5}{6}$

21​

34​

53​

65​

$\frac{1}{2}$

$\frac{4}{3}$

$\frac{3}{5}$

$\frac{5}{6}$