Question

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 $\overrightarrow{E}$ at the centre of curvature O?

• A. $\frac{3\sqrt{3}Q}{8\pi \epsilon_0 R^2}(\hat{i})$
• B. $\frac{3\sqrt{3}Q}{8\pi^2 \epsilon_0 R^2}(\hat{i})$
• C. $\frac{3\sqrt{3}Q}{16\pi^2 \epsilon_0 R^2}(\hat{i})$
• D. $\frac{3\sqrt{3}Q}{8\pi^2 \epsilon_0 R^2}(-\hat{i})$

Answer 1

Answer: (D)

$\frac{3\sqrt{3}Q}{8\pi^2 \epsilon_0 R^2}(-\hat{i})$

Answer 2

Solution Explanation

1. The uniformly-charged 120° arc (subtended angle = 120° = 2π/3 rad) has linear charge density
     λ = –Q/(R·(2π/3)) = –(3Q)/(2πR).
2. Consider a small element at angle θ (measured from the x‑axis). Its charge is
     dq = λR dθ = –(3Q/(2π)) dθ.
3. The contribution to the electric field at the center is
     d𝐸 = (1/(4πε₀))·(dq/R²) (cosθ î + sinθ ĵ).
   Substituting dq gives
     d𝐸 = –(1/(4πε₀))·(3Q/(2πR²)) (cosθ î + sinθ ĵ) dθ.
4. Integrate θ from –π/3 to +π/3. Because the sine (y) components cancel by symmetry, only the x‑component survives:   Eₓ = –(1/(4πε₀))·(3Q/(2πR²)) ∫₋π/3^(π/3) cosθ dθ
       = –(1/(4πε₀))·(3Q/(2πR²))·[sinθ]₋π/3^(π/3)
       = –(1/(4πε₀))·(3Q/(2πR²))·(sin(π/3) – (–sin(π/3)))
       = –(1/(4πε₀))·(3Q/(2πR²))·(2(√3/2))
       = –(3√3Q)/(8π²ε₀ R²).
5. Hence, the net electric field at the center is in the –î direction.