a 80 microfarad capacitor is charged by 50V battery . d capacitor is disconnected from it and connected across uncharged 320 microfarad capacitor calculate the charge on second capacitor

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Solution

$Given, C_{1} = 80 μF = 80 \times 10^{- 6} F = 8 \times 10^{- 5} F V_{1} = 50 V Q_{2} = 0 C_{2} = 320 μF = 320 \times 10^{- 6} F = 3.2 \times 10^{- 4} F Charge on first capacitor = Q_{1} = C_{1} V_{1} = 8 \times 10^{- 5} \times 50 = 4 \times 10^{- 3} C = 4 mC Now, The common potential after connecting them together is given by V = \frac{4 \times 10^{- 3}}{(8 \times 10^{- 5}) + (3.2 \times 10^{- 4})} = 10 V Now, charge on second capacitor will be C_{2} V = 3.2 \times 10^{- 4} \times 10 = 3.2 \times 10^{- 3} F or 3.2 mF or 3200 μF$

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