1. Original capacitance

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

$C_0 = \frac{\varepsilon_0 A}{d}$

2. Effect of inserting a metal sheet

A metal sheet of thickness $\dfrac{d}{2}$ is inserted between the plates.

• Inside a conductor, electric field is zero.
• So, there is no potential drop across the metal sheet.
• The potential drop occurs only across the remaining air gaps.

Hence, the effective separation becomes:

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

3. New capacitance

Thus,

$C' = \frac{\varepsilon_0 A}{d_{\text{eff}}} = \frac{\varepsilon_0 A}{d/2} = \frac{2\varepsilon_0 A}{d}$

So,

$C' = 2C_0$

4. Required ratio

$\frac{C'}{C_0} = 2$

Therefore, the ratio of new capacitance to original capacitance is:

$2:1$

5. Option check

• A: $2:1$ ✅
• B: $1:2$ ❌
• C: $1:4$ ❌
• D: $4:1$ ❌

So, the correct option is A.