Using a battery, a 10 pF capacitor is charged to 50 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is .............. pF.
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Correct Answer: 15
Solution and Explanation
Step 1: Use charge conservation.
The total charge before and after connection must remain constant because the battery is removed.
Initial charge on first capacitor: \[ Q_i = C_1 V_1 = 10 \times 50 = 500 \, \text{pC} \]
Step 2: After connection, voltage becomes common.
Let \(C_2\) be the capacitance of the second capacitor and \(V_f = 20 \, \text{V}\). Final charge on both capacitors combined: \[ Q_f = (C_1 + C_2)V_f \] By conservation of charge: \[ Q_i = Q_f \Rightarrow 500 = (10 + C_2) \times 20 \] \[ 10 + C_2 = 25 \Rightarrow C_2 = 15 \, \text{pF} \]
Step 3: Conclusion.
Hence, the capacitance of the second capacitor = 15 pF.
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