The capacitance of a parallel plate condenser is \( 10\,\mu\mathrm{F} \), when the distance between its plates is \( 8\,\mathrm{cm} \). If the distance between the plates is reduced to \( 4\,\mathrm{cm} \), then its capacitance will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 20\,\mu\mathrm{F} \)
For a parallel plate capacitor, \( C = \varepsilon_0 A/d \). The plate area is unchanged, so the capacitance is inversely proportional to the separation.
\( \displaystyle \frac{C_2}{C_1} = \frac{d_1}{d_2} \)
\( \displaystyle C_2 = 10\,\mu\mathrm{F}\times\frac{8\,\mathrm{cm}}{4\,\mathrm{cm}} \)
\( C_2 = 20\,\mu\mathrm{F} \)
Halving the separation doubles the capacitance.