JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Capacitance

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Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ______ V

Answer: 6

Solution

Initially, the capacitors are in parallel at 18 V:

Charge on capacitor C: $Q_{1} = C \times 18 = 18 C$
Charge on capacitor 3C: $Q_{2} = 3 C \times 18 = 54 C$
Total charge: $Q_{total} = 72 C$

After disconnecting the battery and inserting the dielectric (K = 9) in capacitor C:

New capacitance of first capacitor: $C^{'} = 9 C$
Total capacitance: $C_{total} = 9 C + 3 C = 12 C$

Since charge is conserved:

V^{'} = \frac{Q _{total}}{C _{total}} = \frac{72 C}{12 C} = 6 V

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Capacitance

Solution

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