In a parallel plate capacitor the separation between plates is $3 x$. This separation is filled by two…
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
$\frac{1}{2}$
$\frac{4}{3}$
$\frac{3}{5}$
$\frac{5}{6}$
Solution
Key Idea Capacitance of a parallel plate capacitor is give by relation,
$
C=\varepsilon_r \cdot \frac{\varepsilon_0 A}{d}
$
where, $\varepsilon_r=$ dielectric constant, $A=$ area of parallel plate and $d=$ distance between the plate.
Here, $d_0=3 x, d_1=x, \varepsilon_1=3 k, d_2=2 x$ and $\varepsilon_2=5 k$ The capacitor is shown in the figure below,
Now, $C_1=3 k \frac{\varepsilon_0 A}{x}$ and $C_2=5 k \frac{\varepsilon_0 A}{2 x}$
Since, $m$ in series combination of capacitor, stored charge in each capacitor is same.
So,
$
V_1=\frac{Q}{C_1} \text { and } V_2=\frac{Q}{C_2}
$
$\Rightarrow \quad V_1=\frac{\theta x}{3 k \varepsilon_0 A}$ and $V_2=\frac{\theta 2 x}{5 k \varepsilon_0 A}$
The ratio, $\frac{V_1}{V_2}=\frac{\frac{\theta x}{3 k \varepsilon_0 A}}{\frac{\theta 2 x}{5 k \varepsilon_0 A}}=\frac{5}{3 \times 2}=\frac{5}{6}$
Hence, option (d) is correct
