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A parallel plate capacitor of capacitance 200 μ F is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between the plates of the capacitor while the battery remains connected. The change in the electrostatic energy in the capacitor will be ______ J.

Asked in JEE Main 31st Aug 2nd Shift 2021 · Battery connected or removed

Answer: 4

Step-by-step solution

The battery stays connected, so V is fixed at 200 V and the right energy formula to use is U=1/2CV².

Uᵢ=1/2×200×10⁻⁶×(200)²=4 J.

Inserting the slab doubles the capacitance to 400 μ F, so U_f=2Uᵢ=8 J.

Δ U=8-4=4 J. With the battery connected the energy rises, because the battery supplies the extra charge.

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