A parallel plate capacitor of capacitance \( 2\,\mu\mathrm{F} \) has energy \( 10 \times 10^{-5}\,\mathrm{J} \). The potential difference and the charge on the capacitor are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 10\,\mathrm{V},\ 20\,\mu\mathrm{C} \)
The energy stored in a capacitor is \( U = \frac{1}{2}CV^{2} \), which gives the potential difference; the charge then follows from \( Q = CV \).
\( C = 2\,\mu\mathrm{F} = 2\times10^{-6}\,\mathrm{F} \)
\( U = 10\times10^{-5}\,\mathrm{J} = 10^{-4}\,\mathrm{J} \)
\( \displaystyle V = \sqrt{\frac{2U}{C}} = \sqrt{\frac{2 \times 10^{-4}}{2\times10^{-6}}} = \sqrt{100} = 10\,\mathrm{V} \)
\( \displaystyle Q = CV = 2\times10^{-6} \times 10 = 20\times10^{-6}\,\mathrm{C} = 20\,\mu\mathrm{C} \)