JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Capacitance

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A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 V . The battery is now disconnected and a dielectric slab (

ϵ_{r} = 6

) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________

\times 1 0^{- 12} J

Answer: 750

Solution

Battery is disconnected before the dielectric is inserted, so charge

Q = C_{0} V

stays fixed.
Initial energy:

U_{i} = \frac{Q ^{2}}{2 C _{0}} = \frac{1}{2} C_{0} V^{2} = \frac{1}{2} (12.5) (12)^{2} = 900 pJ

.
After inserting the dielectric,

C_{f} = ϵ_{r} C_{0} = 6 C_{0}

, so

U_{f} = \frac{Q ^{2}}{2 C _{f}} = \frac{U _{i}}{6} = 150 pJ

.
Change in potential energy:

Δ U = U_{i} - U_{f} = 900 - 150 = 750 pJ = 750 \times 1 0^{- 12} J

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Integer answer

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JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Capacitance

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