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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

  1. A0.58
  2. B0.67
  3. C0.82
  4. 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

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