Physics Q38 – JEE-MAIN 2025

Home Physics Electrostatics & CapacitanceQuestion #132

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)

Clue 1 of 3: Charge Conservation

When capacitors are connected in parallel after being isolated, the total charge in the system remains conserved.

🥷

Ninja Strategy•Ratio of Voltages

Since the final voltage ( $20 V$ ) is one-third of the initial voltage ( $60 V$ ), the total capacitance must be three times the initial capacitance of $C_{1}$ ( $300 pF$ ). Only option 2 yields this total capacitance.

🧠 Full Solution Path

Step 1: Calculate Initial Charge✦ Active

The initial charge stored on the first capacitor ( $C_{1} = 100 pF$ ) charged to $V_{1} = 60 V$ is calculated using $Q_{1} = C_{1} V_{1}$ .

Q_{1} = (100 pF) (60 V) = 6000 pC

Step 2: Apply Charge Conservation○ Expand

When the uncharged second capacitor ( $C_{2}$ ) is connected in parallel, the total charge is conserved. The final voltage across both capacitors is $V_{f} = 20 V$ . The total capacitance in parallel is $C_{e q} = C_{1} + C_{2}$ . By charge conservation, the initial total charge equals the final total charge ( $Q_{t o t a l, i n i t i a l} = Q_{t o t a l, f i n a l}$ ).

Q_{1} = (C_{1} + C_{2}) V_{f}

Step 3: Solve for Unknown Capacitance○ Expand

Substitute the known values into the charge conservation equation and solve for $C_{2}$ .

6000 pC = (100 pF + C_{2}) (20 V)

\frac{6000}{20} pF = 100 pF + C_{2}

300 pF = 100 pF + C_{2}

C_{2} = 300 pF - 100 pF = 200 pF

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation.

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Physics Question 38 – JEE-MAIN 2025

Clue 1 of 3: Charge Conservation

🧠 Full Solution Path