A parallel plate capacitor having cross - sectional area 'A' and separated by distance 'd' is filled by copper plate of thickness b. It's capacitance is:

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

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

The correct answer is Option (2) → $\frac{\varepsilon_0 A}{d-b}$