Question

An infinitely long very thin straight wire carries uniform line charge density \( 8\pi \times 10^{-2} \, \text{C/m} \). The magnitude of electric displacement vector at a point located 20 mm away from the axis of the wire is ......... C/m².

Hint:

The electric displacement vector for a line charge is calculated using \( D = \frac{\lambda}{2\pi \epsilon_0 r} \).

Answer 1

Correct Answer: 2

Step 1: Formula for electric displacement vector \( \mathbf{D} \).
The electric displacement vector \( \mathbf{D} \) due to a line charge is given by: \[ D = \frac{\lambda}{2\pi \epsilon_0 r} \] where \( \lambda = 8\pi \times 10^{-2} \, \text{C/m} \) is the charge density, \( r = 20 \, \text{mm} = 0.02 \, \text{m} \), and \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \) is the permittivity of free space.
Step 2: Calculate the electric displacement vector.
Substitute the known values: \[ D = \frac{8\pi \times 10^{-2}}{2\pi \times 8.85 \times 10^{-12} \times 0.02} \] \[ D = \frac{8 \times 10^{-2}}{3.54 \times 10^{-12}} = 2.26 \times 10^{6} \, \text{C/m}^2 \]
Step 3: Conclusion.
The magnitude of the electric displacement vector at a point 20 mm away from the axis of the wire is approximately 1.131 \( \times 10^{-6} \) C/m².