JEE Main 2025 Physics question
A parallel-plate capacitor of capacitance 40μF 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 -
Answer: (3)
Solution
Δq = (K-1)CV = 40×10⁻⁶×100 = 4mC. ΔU = (1/2)(K-1)CV² = 0.2 J.