The separation between the plates of a parallel-plate capacitor is...

Given:
Distance between plates, $d = 0.500 \text{ cm} = 5 \times 10^{-3} \text{ m}$
Area of plates, $A = 100 \text{ cm}^2 = 10^{-2} \text{ m}^2$
Thickness of metal plate, $t = 0.400 \text{ cm} = 4 \times 10^{-3} \text{ m}$

When a metal plate of thickness $t$ is inserted parallel to the plates of a capacitor, it acts as two capacitors in series. Let the metal plate be at a distance $x$ from one plate. The distances of the air gaps are $x$ and $d - t - x$.

The equivalent capacitance $C$ is given by:
$\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$

$\dfrac{1}{C} = \dfrac{x}{\epsilon_0 A} + \dfrac{d - t - x}{\epsilon_0 A} = \dfrac{d - t}{\epsilon_0 A}$

$C = \dfrac{\epsilon_0 A}{d - t}$

Substituting the given values:
$C = \dfrac{8.85 \times 10^{-12} \times 10^{-2}}{5 \times 10^{-3} - 4 \times 10^{-3}}$

$C = \dfrac{8.85 \times 10^{-14}}{10^{-3}}$

$C = 8.85 \times 10^{-11} \text{ F} = 88.5 \text{ pF}$

Since the expression for capacitance does not depend on $x$, the capacitance is independent of the position of the metal plate within the gap.