Question

If in a free space, the electric field is given as: \[ \vec{E} = 20 \cos(\omega t - 50x) \hat{y} \, \text{V/m} \] then, the expression of displacement current density \( J_d \) is:

• A. \( -20 \omega \epsilon_0 \cos(\omega t - 50x) \hat{y} \, \text{A/m}^2 \)
• B. \( -20 \omega \epsilon_0 \sin(\omega t - 50x) \hat{y} \, \text{A/m}^2 \)
• C. \( -10 \omega \epsilon_0 \sin(\omega t - 50x) \hat{y} \, \text{A/m}^2 \)
• D. \( -20 \omega \sin(\omega t - 50x) \hat{y} \, \text{A/m}^2 \)

Hint:

Displacement current density is related to the time rate of change of the electric field in free space.

Answer 1

The Correct Option is B

Step 1: Understanding displacement current.
Displacement current is given by: \[ J_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t} \] Taking the time derivative of the electric field \( \vec{E} = 20 \cos(\omega t - 50x) \hat{y} \), we get: \[ \frac{\partial \vec{E}}{\partial t} = -20 \omega \sin(\omega t - 50x) \hat{y} \] Thus, the displacement current density is: \[ J_d = -20 \omega \epsilon_0 \sin(\omega t - 50x) \hat{y} \]
Step 2: Conclusion.
Thus, the correct expression for displacement current density is option (2).