QJEEM23Ph07fb: JEE Main 2023 Physics - Class 12 JEEM23Ph07fb Solution: The distance between two plates of a capacitor is and its…

When a metal sheet of thickness ( $\frac{2d}{3}$ ) is introduced between the plates of a capacitor, it divides the capacitor into two separate capacitors. The metal sheet acts as a new plate in each capacitor, and because the metal is a conductor, it is at the same potential as the plates on either side.

The capacitance of the original capacitor with air between the plates is given by:

$\mathrm{C}_1 = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}}$

where ( $\epsilon_0$ ) is the permittivity of free space, ( $\mathrm{A}$ ) is the area of one of the plates, and ( $\mathrm{d}$ ) is the distance between the plates.

When the metal sheet of thickness ( $\frac{2d}{3}$ ) is introduced, the distance between the plates is reduced by ( $\frac{2d}{3}$ ), so the capacitance of each of the new capacitors is:

$\mathrm{C}' = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}-\frac{2d}{3}}$

Simplifying, we get:

$\mathrm{C}' = \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}}$

Since the two capacitors are in parallel with each other, the total capacitance is:

$\mathrm{C}_2 = 2\mathrm{C}' = 2 \times \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}} = 3\mathrm{C}_1$

So the ratio of the capacitances is:

$\frac{\mathrm{C}_2}{\mathrm{C}_1} = 3:1$