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QPages 141-142: Q14: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q14 Solution: Find the magnitude of the electric field at a point 4 cm away…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often use the point‑charge field $E= rac{1}{4pivarepsilon_0} rac{Q}{r^2}$ for a line charge, which gives the wrong $1/r^2$ dependence. For an infinite line, the symmetry leads to $E= rac{lambda}{2pivarepsilon_0 r}$, a $1/r$ dependence. Always match the field formula to the charge distribution’s symmetry before plugging in numbers.

Find the magnitude of the electric field at a point 4 cm away from a line charge of density 2×10−6 C m−12 \times 10^{-6} \text{ C m}^{-1}.

Step-by-Step Verified Solution

**Answer:** E=9×105E = 9 \times 10^5 N/C. **Step-by-Step Explanation:** Given: λ=2×10−6\lambda = 2 \times 10^{-6} C/m, r=0.04r = 0.04 m.
E=λ2πϵ0r=2×10−62π×8.85×10−12×0.04=9×105 N/CE = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.04} = 9 \times 10^5 \text{ N/C}

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