Question

The electric field intensity (\(E\)) at a distance of 3 m from a uniform long straight wire of linear charge density 0.2 \(\mu C m^{-1}\) is:

• A. \( 1.2 \times 10^3 \, Vm^{-1} \)
• B. \( 0.6 \times 10^3 \, Vm^{-1} \)
• C. \( 1.8 \times 10^3 \, Vm^{-1} \)
• D. \( 2.4 \times 10^3 \, Vm^{-1} \)

Hint:

For a long charged wire, use the formula \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \). The electric field decreases as the distance \( r \) increases.

Answer 1

The Correct Option is A

Step 1: Apply Electric Field Formula for a Line Charge The electric field due to an infinite line charge is given by: \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \] where: \( \lambda = 0.2 \times 10^{-6} \) C/m (linear charge density), \( r = 3 \) m, \( \varepsilon_0 = 8.85 \times 10^{-12} \, F/m \) (permittivity of free space). Step 2: Calculate the Electric Field \[ E = \frac{(0.2 \times 10^{-6})}{2 \pi (8.85 \times 10^{-12}) (3)} \] \[ E = 1.2 \times 10^3 \, Vm^{-1} \] Thus, the correct answer is \( 1.2 \times 10^3 \, Vm^{-1} \).

Answer 2

Given:
- Linear charge density, \( \lambda = 0.2 \, \mu C/m = 0.2 \times 10^{-6} \, C/m \)
- Distance from the wire, \( r = 3 \, m \)
- We need to find the electric field intensity \( E \) at this distance.

Step 1: Formula for electric field due to a long straight charged wire:
\[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \] where \( \varepsilon_0 = 8.854 \times 10^{-12} \, C^2/N \cdot m^2 \) (permittivity of free space).

Step 2: Substitute the values:
\[ E = \frac{0.2 \times 10^{-6}}{2 \pi \times 8.854 \times 10^{-12} \times 3} \]
Calculate denominator:
\[ 2 \pi \times 8.854 \times 10^{-12} \times 3 = 2 \times 3.1416 \times 8.854 \times 10^{-12} \times 3 \approx 1.67 \times 10^{-10} \]

Step 3: Calculate electric field:
\[ E = \frac{0.2 \times 10^{-6}}{1.67 \times 10^{-10}} \approx 1.2 \times 10^{3} \, V/m \]

Therefore, the electric field intensity at 3 m from the wire is:
\[ \boxed{1.2 \times 10^{3} \, V/m} \]