Q 12-02-091JEE MainJEE Main 2023 (25 Jan, Shift 2)Easy
A capacitor has capacitance $5\ \mu\text{F}$ when its parallel plates are separated by air medium of thickness $d$. A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\dfrac d2$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be ______ $\mu\text{F}$.
Numerical value type. Enter your answer.
Answer: 6
$$C=\frac{\varepsilon_0A}{d-\frac d2+\frac{d}{2\times1.5}}=\frac{\varepsilon_0A}{\frac56d}=\frac65\times5=6\ \mu\text{F}$$
Solution by Sreeraj P, M.Sc Physics