A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

Capacitor PYQ — JEE Main 2003 · Shift 0 · Q166 with Solution

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Correct answer: B

Idea of the situation

A parallel plate capacitor has capacitance $C = \frac{\varepsilon_0 A}{d}$ where:
- $A$ = plate area
- $d$ = separation between plates
We insert a sheet of aluminium foil of negligible thickness between the plates.
What does aluminium do?

Aluminium is a conductor. When a conducting sheet is inserted between the plates, the electric field inside the conductor becomes zero, and the arrangement behaves like two capacitors in series.
Effect of negligible thickness

Suppose the conducting foil has thickness $t$ .

Then the effective gap across which electric field exists becomes: $d - t$ and the new capacitance is $C' = \frac{\varepsilon_0 A}{d-t}$

Since the foil thickness is negligible, we take $t \approx 0$ Hence, $C' = \frac{\varepsilon_0 A}{d} = C$
Conclusion

Therefore, the capacitance remains unchanged.
Checking options
- A: decreases — Incorrect
- B: remains unchanged — Correct
- C: becomes infinite — Incorrect
- D: increases — Incorrect

Therefore, the correct answer is B.