JEE Main20265 April 2026Morning ShiftPhysicsElectromagnetic WavesActual
A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 F capacitor. The rate of change of potential difference across the plates of the capacitor is nearly 10^6 V/s. The value of is __________.
Options
- A0.58
- B0.67
- C0.82
- D0.75
Correct answer
B. 0.67
Step-by-step solution
The displacement current I_d between the plates of a capacitor is given by the formula: I_d = C dV dt Substituting the given values of displacement current and capacitance: 4.0 = 6 10⁻⁶ dV dt dV dt = 4.0 6 10⁻⁶ dV dt = 2 3 10^6 0.67 10^6 V/s Comparing this with the given rate of change of potential difference 10^6 V/s, we get = 0.67 . Answer: 0.67