Question

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.

Hint:

When capacitors are connected after removing the battery, the total charge is conserved while the voltage is redistributed.

Answer 1

Correct Answer: 15

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.