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) → \( C \)
A conducting foil placed parallel to the plates divides the capacitor into two capacitors in series, each with plate area \( A \) and separation \( d/2 \).
\( \displaystyle C = \frac{\varepsilon_0 A}{d} \qquad C_1 = C_2 = \frac{\varepsilon_0 A}{d/2} = 2C \)
\( \displaystyle \frac{1}{C'} = \frac{1}{2C} + \frac{1}{2C} = \frac{1}{C} \;\Rightarrow\; C' = C \)
Because the foil has negligible thickness, it does not reduce the gap between the plates, and the capacitance is unchanged.