Question

A 80uF capacitor is charged by a 50V battery.The capacitor is disconnected from the battery and then connected across another uncharged 320uF capacitor.find the charge on the second capacitor

Answer 1

Answer:

Charge on first capacitor = Q₁ = C₁V₁

                                                = 80u × 50

                                                = 4000u C

                                                = 4m C

Common Potential [After connecting them together]

V = total charge / sum of capacitance

    = (Q₁ + Q₂)/C₁ + C₂ = 4 × 10⁻³/(80+320) × 10⁻⁶ = (4/400) × 10³ = 1000/100 = 10 V

Now, Charge on 2nd capacitor  : = C₂V

                                                    = 320u × 10

                                                    = 3200 u C

                                                    = 3.2m C