The plates of a parallel plate capacitor are separated by a distance d with air as medium between the plates. In order to increase its capacitance by 60%, a slab of dielectric constant 5 having the same area as that of plates is introduced between the plates. The thickness of the dielectric slab is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{15}{32}d \)
With a dielectric slab of thickness \( t \) and dielectric constant \( K \) between the plates, the capacitance is
\( \displaystyle C' = \frac{\varepsilon_0 A}{d - t + t/K} \)
Without the slab, \( C = \varepsilon_0 A/d \). An increase of 60% means \( C' = 1.6\,C \):
\( \displaystyle \frac{C'}{C} = \frac{d}{d - t + t/K} = 1.6 \)
\( \displaystyle d - t + \frac{t}{5} = \frac{d}{1.6} = 0.625\,d \)
\( \displaystyle d - 0.8\,t = 0.625\,d \;\Rightarrow\; 0.8\,t = 0.375\,d \)
\( \displaystyle t = \frac{0.375}{0.8}\,d = \frac{15}{32}\,d \)