JEE Main 2025 Apr 3 Shift 2, Physics Q8: Capacitors And Capacitance

JEE Main2025Apr 3, Shift 2Physics

Using a battery, a $100$ pF capacitor is charged to $60$ 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 : (in pF)

A
$600$
B
$200$
C
$400$
D
$100$

Solution

Charge is conserved after the battery is removed. Initial charge on $C_{0} = 100$ pF is $Q_{0} = C_{0} V_{0} = 100 \cdot 60 = 6000$ pC.

After paralleling with an unknown $C$ , both share the same final voltage $V_{f} = 20$ V:

$Q_{0} = (C_{0} + C) V_{f} \Rightarrow 6000 = (100 + C) \cdot 20 \Rightarrow 100 + C = 300 \Rightarrow C = 200$ pF.

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JEE Main 2025 Apr 3 Shift 2, Physics Q8: Capacitors And Capacitance

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