Verified Solution • 2025–26 Session moderate

QPages 121-124: Q44: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q44 Solution: A circular wire-loop of radius a carries a total charge Q…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often forget that the complete ring produces zero field at its centre, so the field after cutting a piece is the *negative* of the field that the removed piece would have generated; treating it as a positive contribution flips the direction and gives the wrong sign.

A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the centre due to the remaining wire.

Step-by-Step Verified Solution

**Answer:**

E = \frac{kQ dL}{2\pi a^3}

. **Step-BY-STEP Explanation:** When a small length

dL

is removed, the remaining charge distribution is equivalent to the original ring plus a negative charge

-dq = -\frac{Q}{2\pi a}dL

at the removed position. The field due to the remaining wire at the centre equals the field due to this negative charge:

E = \frac{k dq}{a^2} = \frac{kQ dL}{2\pi a^3}

All Questions in this Chapter

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