A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacitance is:
Answer & explanation
Correct answer: option 3
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 -
$C=\frac{ε_0A}{d_{effective}}=\frac{\varepsilon_0 A}{d-b}$