A parallel-plate air capacitor has a plate area of \( 100\,\mathrm{cm^2} \) and a plate separation of 5 mm. A potential difference of 300 V is established between its plates by a battery. After disconnecting the battery, the space between the plates is completely filled with a dielectric of dielectric constant K = 2.5. The new potential difference between the plates of the capacitor is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 120 V
Once the battery is disconnected, the charge Q on the plates stays constant. Filling the gap with a dielectric raises the capacitance from \( C_0 \) to \( KC_0 \), so the potential difference \( V=Q/C \) falls to \( 1/K \) of its value. The plate area and separation are not needed.
\( \begin{aligned} V' &= \frac{Q}{KC_0} \\ &= \frac{V_0}{K} \\ &= \frac{300\,\mathrm{V}}{2.5} \\ &= 120\,\mathrm{V} \end{aligned} \)