Δq=(KC−C)V=40×10−6×100=4000×10−3=4mCΔU=21C′V2−21CV2=21( K−1)CV2=21CV2(2−1)=21CV2=21×40×10−6×10000=0.2 J
A parallel-plate capacitor of capacitance 40 μ ~F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with…
JEE Main 2025 — Physics Electromagnetism
A parallel-plate capacitor of capacitance 40μ 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=2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are
Held on 22 Jan 2025 · Verified 6 Jul 2026.
Options
- A
4 mC and 0.2 J
- B
8 mC and 2.0 J
- C
2 mC and 0.4 J
- D
2 mC and 0.2 J
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