Physics-Electric Field

NEET Physics Electric Field MCQ Question

Type: MCQ-numerical-Hard-Class 12

For a long straight charged wire with linear charge density λ=5×106C/m\lambda = 5 \times 10^{-6} \text{C/m}, what is the electric field EE at a distance r=0.2mr = 0.2 \text{m} from the wire? Assume ε0=8.85×1012C2/Nm2\varepsilon_0 = 8.85 \times 10^{-12} \text{C}^2/\text{N} \cdot \text{m}^2.

A

4.5×104N/C4.5 \times 10^4 \text{N/C}

B

3.0×104N/C3.0 \times 10^4 \text{N/C}

C

9.0×104N/C9.0 \times 10^4 \text{N/C}

D

2.2×104N/C2.2 \times 10^4 \text{N/C}

Correct Answer

Option A

Detailed Explanation

Using Gauss's law, the electric field EE for a long straight wire is given by E=λ2πε0rE = \frac{\lambda}{2\pi\varepsilon_0 r}. Substituting the given values, E=5×1062π(8.85×1012)(0.2)=4.5×104N/CE = \frac{5 \times 10^{-6}}{2\pi(8.85 \times 10^{-12})(0.2)} = 4.5 \times 10^4 \text{N/C}.

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