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. $\frac{1}{2}$
• B. $\frac{4}{3}$
• C. $\frac{3}{5}$
• D. $\frac{5}{6}$

Answer 1

Answer: (D)

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, $V_{1}=\frac{Q}{C_{1}}$ and $V_{2}=\frac{Q}{C_{2}}$
$\Rightarrow V_{1}=\frac{\theta x}{3 k \varepsilon_{0} A}$ and
$V_{2}=\frac{\theta 2 x}{5 k \varepsilon_{0} A}$
The ratio, $\frac{V_{1}}{V_{2}}=\frac{\frac{\theta x}{3 k \varepsilon_{0} A}}{\frac{\theta 2 x}{5 k \varepsilon_{0} A}}$
$=\frac{5}{3 \times 2}=\frac{5}{6}$