The distance between two plates of a capacitor is $$d$$ and its capacitance is $$C_1$$, when air is the medium between the plates. If a metal sheet of thickness $$\frac{2d}{3}$$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $$C_2$$. The ratio $$\frac{C_2}{C_1}$$ is

We need to find the ratio $$\frac{C_2}{C_1}$$ when a metal sheet of thickness $$\frac{2d}{3}$$ is introduced between the plates of a parallel plate capacitor with plate separation $$d$$.

We begin by noting that for a parallel plate capacitor with air as the medium, the capacitance without the metal sheet is given by $$C_1 = \frac{\varepsilon_0 A}{d}$$, where $$\varepsilon_0$$ is the permittivity of free space, $$A$$ is the area of the plates, and $$d$$ is the distance between the plates.

Next, we recall that a metal (conductor) is a perfect conductor, which means the electric field inside the metal is zero. When a metal sheet of thickness $$t$$ is placed between the plates of a capacitor, it effectively reduces the gap over which the electric field exists. The metal sheet acts as an equipotential region with no field inside it, so the effective distance over which the electric field acts is the total gap minus the thickness of the metal sheet.

This effective distance is given by
$$d_{eff} = d - t$$.

Substituting $$t = \frac{2d}{3}$$ into this expression yields
$$d_{eff} = d - \frac{2d}{3} = \frac{3d - 2d}{3} = \frac{d}{3}$$.

Therefore, the new capacitance becomes
$$C_2 = \frac{\varepsilon_0 A}{d_{eff}} = \frac{\varepsilon_0 A}{d/3} = \frac{3\varepsilon_0 A}{d} = 3C_1$$.

Finally, taking the ratio yields
$$\frac{C_2}{C_1} = \frac{3C_1}{C_1} = 3$$, so that $$C_2 : C_1 = 3 : 1$$.

Hence, the correct answer is Option 1: 3 : 1.