Case: Read the passage and answer the question.
In a parallel plate air capacitor having plate separation 5 cm, an electric field of \( 4\times10^{4}\,\mathrm{V/m} \) is established between the plates. After the removal of the battery, a metal plate of thickness t = 2 cm is inserted between the plates of the capacitor.
The potential difference across the capacitor is measured while the metal plate is pushed slowly into the gap. How does the potential difference change?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → It decreases.
With the battery removed, the charge on the capacitor plates cannot change. So the field in the air between the plates stays the same. The potential difference is this field multiplied by the length of air it acts over.
Option (1) It decreases. — Correct. The field inside the metal is zero, so the part of the gap filled by metal adds nothing to the potential difference. As the plate goes in, a smaller length of air carries the field. The potential difference therefore falls.
Option (2) It increases. — Incorrect. A rise would need a stronger field in the air. The field stays the same because the charge on the plates is fixed.
Option (3) It stays the same. — Incorrect. The field in the air is unchanged, but the length of air it acts over becomes shorter. So the potential difference does not stay the same.
Option (4) It drops to zero as soon as the plate enters. — Incorrect. The metal plate does not touch the capacitor plates. Air gaps of total width \( d-t \) remain, so some potential difference remains.
NCERT (2026-27):
Electrostatics of Conductors – “Electrostatic field is zero inside a conductor.”
The Parallel Plate Capacitor – “Now for uniform electric field, potential difference is simply the electric field times the distance between the plates, that is,”
\( \displaystyle V=Ed=\frac{1}{\varepsilon_0}\frac{Qd}{A} \qquad (2.42) \)