A parallel plate capacitor with plates of area A, held a distance d apart, has a capacitance C. A thin silver foil of negligible thickness and the same area is inserted at mid-point between the plates. The capacitance becomes
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → C
Given:
Original capacitance, $C = \frac{\varepsilon_0 A}{d}$
When a thin conducting silver foil is inserted midway between the plates, it divides the capacitor into two capacitors in series, each with plate separation $\frac{d}{2}$.
Capacitance of each section:
$C_1 = C_2 = \frac{\varepsilon_0 A}{(d/2)} = \frac{2\varepsilon_0 A}{d} = 2C$
Since they are in series:
$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2C} + \frac{1}{2C} = \frac{1}{C}$
⟹ $C' = C$
Final Answer: The capacitance remains unchanged, $C' = C$