Verified Solution • 2025–26 Session moderate

QPages 121-124: Q40: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q40 Solution: A rod of length L has a total charge Q distributed uniformly…

⚠️ Common Exam Traps & Conceptual Pitfalls

When replacing $R$ by $L/\pi$ in an expression that already contains a factor $\pi R^2$, many students cancel $\pi$ incorrectly and lose the extra $\pi$ factor, ending up with $E=2kQ/L^2$ instead of the correct $E=2\pi kQ/L^2$. Always substitute $R=L/\pi$ first, then simplify $\pi R^2 = \pi (L^2/\pi^2)=L^2/\pi$, so the final result retains a factor $\pi$ in the numerator.

A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.

Step-by-Step Verified Solution

**Answer:**

E = \frac{2kQ}{L^2}

. **Step-BY-STEP Explanation:** For a semicircular rod of length

L = \pi R

, the electric field at the centre is

E = \frac{2kQ}{\pi R^2}

. Since

R = L/\pi

,

E = \frac{2kQ}{L^2}

.

All Questions in this Chapter

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