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JEE Main 2025 (22 Jan, Shift 1) · Dielectrics

Dielectrics question from JEE Main 2025 (22 Jan, Shift 1)

Q32JEE Main 2025 · 22 JAN · SHIFT 1Medium

Dielectrics

A parallel-plate capacitor of capacitance 40μ F40\mu\text{ F} is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K=2K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are

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Answer: (b)

  1. Supply stays connected, so V=100V = 100 V is fixed and C′=KC=80 μFC' = KC = 80\ \mu\text{F}.
  2. Extra charge =(C′−C)V=40×10−6×100=4×10−3= (C' - C)V = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3} C =4= 4 mC.
  3. Change in energy =12(C′−C)V2=12×40×10−6×104=0.2= \frac{1}{2}(C' - C)V^2 = \frac{1}{2} \times 40 \times 10^{-6} \times 10^4 = 0.2 J.
  4. So the answer is (b) 4 mC and 0.2 J.
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