Q 12-02-111JEE MainJEE Main 2022 (25 Jun, Shift 2)Easy
Two metallic plates form a parallel plate capacitor. The distance between the plates is $d$. A metal sheet of thickness $\dfrac d2$ and of area equal to the area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?
Answer: (B) $2 : 1$
A metal sheet of thickness $t$ reduces the effective gap to $d - t$:
$$C' = \frac{\varepsilon_0A}{d - d/2} = \frac{2\varepsilon_0A}{d} = 2C$$
Ratio $= 2 : 1$.
Solution by Sreeraj P, M.Sc Physics