Question

A 80 mf capacitor i charged by a 50 v battery the capacitor us disconnected from the battery and then connected across another uncharged 320 mf capacitor. calculate charge on the second capacitor

Answer 1

Heya ,

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Q1 = charge on first capacitor = C1 V1 = 80u x 50 = 4000u C = 4mC

common potential after connecting them together

V = total charge / sum of capacitances

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

charge on second capacitor = C2 V = 320u x 10 = 3200u C = 3.2 m C


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Hope it helps u !!!

# Nikky

Answer 2

Hey there!

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 

Hope It Helps You!