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.
When the metal plate is placed between the plates, which charge appears on the face of the metal plate nearer to the positive plate of the capacitor?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A negative charge, equal in size to the charge on the positive plate
The metal plate has free electrons. The field between the capacitor plates makes these electrons drift against the field, that is, towards the positive plate.
Option (1) A positive charge, equal to the charge on the positive plate — Incorrect. The electrons gather on the face near the positive plate. That face becomes negative.
Option (2) A negative charge, equal in size to the charge on the positive plate — Correct. The electrons keep moving until the field of the induced charges is equal and opposite to the outside field within the metal. For this, the induced charge on the near face must be equal in size to the charge on the positive plate. The far face is left with an equal positive charge.
Option (3) A negative charge, smaller in size than the charge on the positive plate — Incorrect. A smaller induced charge would leave some field inside the metal. In the static situation, a conductor has no field inside it.
Option (4) No charge, because the metal plate as a whole is neutral — Incorrect. The plate does stay neutral as a whole. Its two faces still get equal and opposite charges.
NCERT (2026-27):
Electrostatics of Conductors – “In an external electric field, they drift against the direction of the field.”
Dielectrics and Polarisation – “The free charge carriers move and charge distribution in the conductor adjusts itself in such a way that the electric field due to induced charges opposes the external field within the conductor. This happens until, in the static situation, the two fields cancel each other and the net electrostatic field in the conductor is zero.”