A capacitor of unknown capacitance is connected with a battery of V volts. The charge stored in it is 200 μC. When the potential of the capacitor is reduced by 100 volts, the charge stored in it becomes 150 μC. The unknown capacitance is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.5 μF
For a capacitor \( Q = CV \) with \( C \) fixed, so a change in charge is proportional to the change in potential: \( \Delta Q = C\,\Delta V \).
\( \displaystyle C = \frac{\Delta Q}{\Delta V} = \frac{(200 - 150)\,\mu\mathrm{C}}{100\,\mathrm{V}} = \frac{50\,\mu\mathrm{C}}{100\,\mathrm{V}} = 0.5\,\mu\mathrm{F} \)
Check: \( V = 200\,\mu\mathrm{C}/0.5\,\mu\mathrm{F} = 400\,\mathrm{V} \), and at \( 300\,\mathrm{V} \) the charge is \( 150\,\mu\mathrm{C} \), as given.