A parallel plate air capacitor is charged to a potential difference of 120 V and then disconnected from the battery. A metal plate of the same area as the plates and of thickness one-fourth of the plate separation is then placed between the plates, parallel to them. The potential difference between the plates now becomes
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 90 V
Once the battery is disconnected, the charge on the plates stays fixed. The field inside the metal plate is zero. So only the two air gaps, of total width \( d-t \), give a potential difference. The capacitor then behaves as an air capacitor of separation \( d-t \).
With \( t=d/4 \), the new capacitance is:
\( \begin{aligned} C' &= \frac{\varepsilon_0 A}{d-d/4} \\ &= \frac{4}{3}\,\frac{\varepsilon_0 A}{d} \\ &= \frac{4C}{3} \end{aligned} \)
For the same charge \( Q=CV \), the potential difference becomes:
\( \begin{aligned} V' &= \frac{Q}{C'} \\ &= \frac{3}{4}\,V \\ &= \frac{3}{4}\times120\,\mathrm{V} \\ &= 90\,\mathrm{V} \end{aligned} \)
NCERT (2026-27):
Electrostatics of Conductors – “Consider a conductor, neutral or charged. There may also be an external electrostatic field. In the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor. This fact can be taken as the defining property of a conductor. …”