The capacitance of a capacitor becomes \( \displaystyle\frac{7}{6} \) times its original value if a dielectric slab of thickness \( t = \displaystyle\frac{2}{3}d \) is introduced in between the plates, where d is the separation between the plates. The dielectric constant of the slab is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{14}{11} \)
A dielectric slab of thickness \( t \) and dielectric constant \( K \) inside a parallel plate capacitor reduces the field in the slab by \( K \), so the capacitance becomes
\( \displaystyle C' = \frac{\varepsilon_0 A}{d - t + t/K} \)
Since \( C = \varepsilon_0 A/d \), the condition \( C' = \frac{7}{6}C \) gives
\( \displaystyle d - t + \frac{t}{K} = \frac{6}{7}\,d \)
Putting \( t = \frac{2}{3}d \):
\( \displaystyle \frac{d}{3} + \frac{2d}{3K} = \frac{6d}{7} \)
\( \displaystyle \frac{2}{3K} = \frac{6}{7} - \frac{1}{3} = \frac{18 - 7}{21} = \frac{11}{21} \)
\( \displaystyle K = \frac{2 \times 21}{3 \times 11} = \frac{14}{11} \)
A dielectric constant must be greater than 1, and \( 14/11 \approx 1.27 \) is.