1. Initial capacitance with air

For a parallel plate capacitor with plate area $A$ and separation $d$:

$C_1 = \frac{\varepsilon_0 A}{d}$

2. When a metal sheet is inserted

A metal sheet of thickness $\frac{2d}{3}$ is placed between the plates.

Inside a conductor, electric field is zero, so there is no potential drop across the metal sheet. Hence the effective separation across which potential drops occur is only the remaining air gap.

So effective air separation becomes:

$d_{\text{eff}} = d - \frac{2d}{3} = \frac{d}{3}$

Thus the new capacitance is:

$C_2 = \frac{\varepsilon_0 A}{d/3} = \frac{3\varepsilon_0 A}{d}$

3. Ratio

$\frac{C_2}{C_1} = \frac{\frac{3\varepsilon_0 A}{d}}{\frac{\varepsilon_0 A}{d}} = 3$

So,

$\frac{C_2}{C_1} = 3:1$

4. Option check

• A: $1:1$ ✗
• B: $3:1$ ✓
• C: $2:1$ ✗
• D: $4:1$ ✗

Therefore, the correct option is B.