Answer to Question #207120 in Physics for Favorite

Question

Two capacitors of capacitance 3uF and 4uF respectively, are connected in parallel with batter of emf of 220volts. The connections are broken and the like terminals of the capacitors are joined. Find the final charge on each capacitor (L)

Answer 1

$$Q=(C_1+C_2)U\\=(3+4)10^{-6}\cdot220=1.54\cdot10^{-3}\ C\\\frac{q_1}{3\cdot10^{-6}}=\frac{1.54\cdot10^{-3}-q_1}{4\cdot10^{-6}}\\q_1=6.6\cdot10^{-4}C\\q_2=8.8\cdot10^{-4}C$$