The capacitance of a capacitor becomes $\frac{9}{7}$ times of its original value when a dielectric slab of thickness $t =\frac{d}{4}$ is inserted 7 between the plates, where d is the separation between the plates. The dielectric constant of the slab is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9
Original capacitance: $C_0 = \frac{\varepsilon_0 A}{d}$
Dielectric slab of thickness $t = \frac{d}{4}$ and dielectric constant $K$ is inserted between plates. This forms two capacitors in series:
Capacitance of slab: $C_1 = \frac{K \varepsilon_0 A}{t} = \frac{K \varepsilon_0 A}{d/4} = \frac{4 K \varepsilon_0 A}{d}$
Capacitance of remaining air gap: $C_2 = \frac{\varepsilon_0 A}{d - t} = \frac{\varepsilon_0 A}{d - d/4} = \frac{\varepsilon_0 A}{3d/4} = \frac{4 \varepsilon_0 A}{3d}$
Total capacitance (series):
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{4 K \varepsilon_0 A} + \frac{3 d}{4 \varepsilon_0 A} = \frac{d}{4 \varepsilon_0 A} \left( \frac{1}{K} + 3 \right)$
$C = \frac{4 \varepsilon_0 A}{d \left( 3 + \frac{1}{K} \right)} = \frac{4 C_0}{3 + \frac{1}{K}}$
Given $C = \frac{9}{7} C_0$
$\frac{4 C_0}{3 + \frac{1}{K}} = \frac{9}{7} C_0 \Rightarrow 3 + \frac{1}{K} = \frac{4 \cdot 7}{9} = \frac{28}{9}$
$\frac{1}{K} = \frac{28}{9} - 3 = \frac{28}{9} - \frac{27}{9} = \frac{1}{9}$
$K = 9$
Answer: Dielectric constant $K = 9$