A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacitance is:

$\frac{\varepsilon_0 A}{d+2 b}$ $\frac{\varepsilon_0 A}{2 d}$ $\frac{\varepsilon_0 A}{d-b}$ $\frac{2 \varepsilon_0 A}{d+2 b}$

$\frac{\varepsilon_0 A}{d-b}$

The correct answer is Option (3) → $\frac{\varepsilon_0 A}{d-b}$