The capacitance of a capacitor becomes \( \displaystyle\frac{9}{7} \) times of its original value when a dielectric slab of thickness \( t = \displaystyle\frac{d}{4} \) is inserted 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
A dielectric slab of thickness \( t \) and dielectric constant \( K \) inside a parallel plate capacitor acts as if the air gap were reduced from \( d \) to \( d - t + t/K \), so it increases the capacitance.
\( \displaystyle C = \frac{\varepsilon_0 A}{d} \qquad C' = \frac{\varepsilon_0 A}{d - t + t/K} \)
\( \displaystyle \frac{C'}{C} = \frac{d}{d - t\left(1 - \frac{1}{K}\right)} = \frac{9}{7} \)
Putting \( t = d/4 \):
\( \displaystyle d - \frac{d}{4}\left(1 - \frac{1}{K}\right) = \frac{7d}{9} \)
\( \displaystyle \frac{1}{4}\left(1 - \frac{1}{K}\right) = 1 - \frac{7}{9} = \frac{2}{9} \;\Rightarrow\; 1 - \frac{1}{K} = \frac{8}{9} \)
\( \displaystyle \frac{1}{K} = \frac{1}{9} \;\Rightarrow\; K = 9 \)