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

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

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

The effective seperation between the plates is -

$d_{effective}=d-b$

Now,

The capacitance of a parallel plate capacitor is -

$C=\frac{ε_0A}{d_{effective}}=\frac{\varepsilon_0 A}{d-b}$