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A 10 μ F 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. The potential difference across them now becomes 20 V. The capacitance of the second capacitor is

Asked in JEE Main 2nd Sept 2nd Shift 2020 · Redistribution of charge

• (1) 15 μ F
• (2) 20 μ F
• (3) 30 μ F
• (4) 10 μ F

Step-by-step solution

Why the other options are wrong

• (2) Doubles the first capacitance because the voltage fell by more than half; the total would then be 30 μ F and the common potential 16.7 V, not 20 V.
• (3) Triples the first capacitance; the total would be 40 μ F and the common potential only 12.5 V.
• (4) Two identical capacitors would simply halve the potential to 25 V, not drop it to 20 V.

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