Question

A 80×10^-12F capacitor is charged by a 50volt battery.the capacitor is disconnected from the battery and connecred across another uncharged 320×10^-12F capacitor.calculate the charge and the energy stored on the secound capacitor

Answer 1

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