Question

Which of the following represent correct form of displacement current?

• A. \(\dfrac{1}{2}\epsilon_0E^2\)
• B. \(\dfrac{1}{2}\dfrac{B^2}{\mu_0}\)
• C. \(\vec{E}\times\vec{B}\)
• D. \(\epsilon_0\dfrac{\partial \vec{E}}{\partial t}\)

Hint:

Displacement current density is produced by time-varying electric field and is given by \(\vec{J}_d=\epsilon_0\frac{\partial\vec{E}}{\partial t}\).

Answer 1

The Correct Option is D

Concept:
Displacement current density was introduced by Maxwell to modify Ampere's law for time-varying electric fields.

Step 1: Displacement current density.
The displacement current density is: \[ \vec{J}_d=\epsilon_0\frac{\partial \vec{E}}{\partial t} \]

Step 2: Check other options.
\[ \frac{1}{2}\epsilon_0E^2 \] is electric energy density. \[ \frac{1}{2}\frac{B^2}{\mu_0} \] is magnetic energy density. \[ \vec{E}\times\vec{B} \] is related to electromagnetic energy flow.

Step 3: Final answer.
Thus, the correct form of displacement current density is: \[ \epsilon_0\frac{\partial \vec{E}}{\partial t} \] \[ \therefore \text{Correct Answer is (D)} \]