Two capacitors of capacitances \( 3\,\mu\mathrm{F} \) and \( 6\,\mu\mathrm{F} \) are charged to potentials of 2 V and 3 V, respectively. Now, if they are connected in parallel, the charge on each of the two capacitors will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 8\,\mu\mathrm{C} \) and \( 16\,\mu\mathrm{C} \)
When the capacitors are joined in parallel (positive plate to positive plate), the total charge is conserved and they settle at a common potential.
\( Q_1 = C_1V_1 = 3 \times 2 = 6\,\mu\mathrm{C} \)
\( Q_2 = C_2V_2 = 6 \times 3 = 18\,\mu\mathrm{C} \)
\( \displaystyle Q = Q_1 + Q_2 = 24\,\mu\mathrm{C} \qquad C = C_1 + C_2 = 9\,\mu\mathrm{F} \)
\( \displaystyle V = \frac{Q}{C} = \frac{24}{9} = \frac{8}{3}\,\mathrm{V} \)
\( \displaystyle Q_1' = 3 \times \frac{8}{3} = 8\,\mu\mathrm{C} \)
\( \displaystyle Q_2' = 6 \times \frac{8}{3} = 16\,\mu\mathrm{C} \)