Question

A 80micro farad capacitor is charged by a 50v battery.the capacitor is disconnected from the battery and then connected across uncharged 320microFarad capacitor.calculate the charge and energy stored on the second 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 

Answer 2

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  

Explanation: