A 10 μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :

Solution

Physics sir · 3 steps

The battery is gone, so the charge stays fixed. Share it between the two capacitors at the common 20 V.

1. Charge at start. The first capacitor holds Q = CV = 10 μF × 50 V = 500 μC. With the source removed, this charge has nowhere to go except onto the second capacitor.
2. Charge is conserved. In parallel, both capacitors end up at the same potential difference, 20 V. The total charge on them is still 500 μC.

   (10 + C₂) × 20 = 500

3. So. 10 + C₂ = 25, which gives C₂ = 15 μF, option (b). Check: the first capacitor keeps 200 μC and the second takes 300 μC, which adds up to 500 μC.

The idea to keep. Once the battery is disconnected, total charge is conserved. In parallel the common voltage is total charge divided by total capacitance: V = C₁V₁/(C₁ + C₂).

Answer from the NTA final answer key. Solution by PrepWiser.