AIIMS 2018 Physics Capacitance MCQ Question

When a dielectric slab with a dielectric constant (κ) of 5 is introduced into a parallel plate capacitor, the effective capacitance increases according to the formula $C' = C \cdot \frac{1}{1 - \frac{t}{d} + \frac{t}{d} \cdot \kappa}$. Given that the original capacitance $C$ is 100 μF, and since $t << d$, the capacitance effectively becomes $C' \approx C \cdot \kappa = 100 \, \mu\text{F} \cdot 5 = 500 \, \mu\text{F}$.

Options A (50 μF) and B (100 μF) are incorrect as they do not account for the increase in capacitance due to the dielectric, while option D (500 μF) is the correct calculation based on the introduction of the dielectric slab.