Verified Solution • 2025–26 Session moderate

QPages 121-124: Q41: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q41 Solution: A 10-cm long rod carries a charge of distributed uniformly…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly calculate the field by treating the rod as a point charge or by summing magnitudes without accounting for vector components. You must resolve the electric field contributions into components, recognizing that the perpendicular components cancel out due to symmetry, leaving only the component along the axis of the rod.

A 10-cm long rod carries a charge of +50μC+50 \mu\text{C} distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both ends of the rod.

Step-by-Step Verified Solution

**Answer:** E=1.35×107E = 1.35 \times 10^7 N/C. **Step-BY-STEP Explanation:** Given: Q=50×10−6Q = 50 \times 10^{-6} C, L=10L = 10 cm, point at 10 cm from both ends (equilateral triangle). Using the formula for a finite line charge:
E=kQxx2+L2/4×2sin⁡θE = \frac{kQ}{x\sqrt{x^2 + L^2/4}} \times 2\sin\theta
Standard answer: 1.35×1071.35 \times 10^7 N/C .

All Questions in this Chapter

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