Figure shows a rod AB, which is bent in a 120° circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric field E⃗ at the centre of curvature O ?

Solution

Physics sir · 4 steps

Symmetry cancels every y-part, so you only need to add up the x-parts from each small piece of the arc.

1. Charge per length. The arc spans 120° = 2π/3 rad, so its length is (2π R)/3. The charge per unit length has size λ = Q/(2π R/3) = 3Q/(2π R).
2. Use symmetry. Take a small piece at angle φ from the x-axis, with φ running from -60° to +60°. Its field at O has size (kλ R dφ)/R². The pieces at +φ and -φ cancel each other's y-parts, so only the x-parts, with factor cos φ, add up.

   E = ∫_(-π/3)^(π/3) kλ/R cos φ dφ = kλ/R(2 sin 60°) = (√3 kλ)/R

3. Put in numbers. E = √3 × 1/4πε₀ × 3Q/(2π R) × 1/R = (3√3 Q)/(8π²ε₀ R²).
4. So. The charge is negative, so the field at O points towards the arc, which lies on the +x side. That gives (3√3 Q)/(8π²ε₀ R²)(î), option (b).

The idea to keep. For an arc of angle θ at its centre, E = 2kλ/R sin θ/2, along the arc's axis of symmetry. A negative charge pulls the field towards the arc, a positive one pushes it away.

Answer from the NTA final answer key. Solution by PrepWiser.