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) → \( \displaystyle\frac{\varepsilon_0 A}{d - b} \)
Copper is a conductor, and the electric field inside a conductor is zero. So the field \( E_0 = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A) \) exists only in the air gap of total width \( d - b \).
\( \displaystyle V = E_0\,(d - b) = \frac{Q\,(d - b)}{\varepsilon_0 A} \)
\( \displaystyle C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d - b} \)
The copper plate shortens the effective separation from \( d \) to \( d - b \), so the capacitance increases.
Note: A similar question was also asked in CUET (UG) 2022 on 18-08-2022 (Slot 2).