A metal foil of neglible thickness is introduced between two parallel plates of a capacitor at the centre. The new capacitance of the capacitor would be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → C
The metal foil becomes an equipotential surface with induced charges on its faces, so it splits the capacitor into two capacitors in series, each with plates a distance \( d/2 \) apart. For a parallel plate capacitor \( C = \varepsilon_0A/d \), so halving the gap doubles the capacitance.
\( \displaystyle C_1 = C_2 = \frac{\varepsilon_0A}{d/2} = 2C \)
\( \displaystyle \frac{1}{C_{\mathrm{new}}} = \frac{1}{2C} + \frac{1}{2C} = \frac{1}{C} \)
\( C_{\mathrm{new}} = C \)
Shortcut: A conducting slab of thickness t gives \( C = \varepsilon_0A/(d - t) \), wherever it is placed between the plates. For a foil of negligible thickness, \( t\to0 \) and the capacitance stays C.
Note: A similar question was also asked in CUET (UG) 2026 on 18-05-2026 (Slot 2), and in CUET (UG) 2023 on 11-06-2023 (Slot 3).