The distance between two plates of a capacitor is $\mathrm{d}$ and its capacitance is $\mathrm{C}_{1}$, when air is the medium between the plates. If a metal sheet of thickness $\frac{2 d}{3}$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $\mathrm{C}_{2}$. The ratio $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$ is

<p>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. </p> <p>The capacitance of the original capacitor with air between the plates is given by:</p> <p>$\mathrm{C}_1 = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}}$</p> <p>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.</p> <p>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:</p> <p>$\mathrm{C}&#39; = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}-\frac{2d}{3}}$</p> <p>Simplifying, we get:</p> <p>$\mathrm{C}&#39; = \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}}$</p> <p>Since the two capacitors are in parallel with each other, the total capacitance is:</p> <p>$\mathrm{C}_2 = 2\mathrm{C}&#39; = 2 \times \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}} = 3\mathrm{C}_1$</p> <p>So the ratio of the capacitances is:</p> <p>$\frac{\mathrm{C}_2}{\mathrm{C}_1} = 3:1$</p>