Physics · Electromagnetic Waves

JEE Main 2026 — 5 April, Morning Shift — Question 11

A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 μF capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $\alpha \times 10^6$ V/s. The value of $\alpha$ is ______.

Option A:
0.58
Option B:
0.67
Correct
Option C:
0.82
Option D:
0.75

Answer: B

Step-by-step solution

$I_d = C \frac{dV}{dt}$ ⇒ $4 = 6\times10^{-6} \times \frac{dV}{dt}$ ⇒ $\frac{dV}{dt} = \frac{4}{6}\times10^6 = 0.6667\times10^6$ V/s, so $\alpha = 0.67$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Displacement Current and Equation of EM Waves