Two capacitors of capacitances \( 2\,\mu\mathrm{F} \) and \( 4\,\mu\mathrm{F} \) are connected in parallel. A charge of \( 6\,\mu\mathrm{C} \) is given to the combination which gets shared. The ratio of the charges on the two capacitors, respectively will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 : 2
Capacitors in parallel have the same potential difference across them, so the charge on each is proportional to its capacitance.
\( \displaystyle V = \frac{Q}{C_1 + C_2} = \frac{6\,\mu\mathrm{C}}{6\,\mu\mathrm{F}} = 1\,\mathrm{V} \)
\( Q_1 = C_1 V = 2\,\mu\mathrm{C} \)
\( Q_2 = C_2 V = 4\,\mu\mathrm{C} \)
\( \displaystyle \frac{Q_1}{Q_2} = \frac{C_1}{C_2} = \frac{2}{4} = 1 : 2 \)