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Field at center of bent charged rod

Short answer

Linear charge density λ = 3Q/(2πr). By symmetry only the x‑component survives: E = (λ/4πε₀r) ∫ cosθ dθ from −60° to 60° = (λ√3)/(4πε₀r). Substituting λ gives E = (3√3/(2π)) Q/(4πε₀r²) ≈ 0.83 Q/(4πε₀r²).

A plastic rod contains uniformly distributed Q charge. The rod has been bent in a 120° circular arc of radius 'r' as shown in the figure [Figure: a circular arc of 120° subtending its centre, radius r, centre at the origin, arc symmetric about the horizontal axis]. Prove that the electric field intensity at the center of the bent rod is E = 0.83Q/(4πε0r²).

Let the arc of radius r subtend 120° at the centre O (take Q positive), symmetric about the x-axis, running from θ = −60° to θ = +60°.

          y
          |    . 
          |  .  (arc, 120 deg)
   -------O-------)-----> x
          |  .   
          |    .
   each end at 60 deg from the x-axis

Steps

  1. Arc length L = r × (2π/3) = 2πr/3. Linear charge density λ = Q/L = 3Q/(2πr).
  2. Take an element dl = r dθ at angle θ, carrying dq = λ r dθ.
  3. Its field at O has magnitude dE = (1/4πε0) dq/r² = (1/4πε0) λ dθ / r. For positive Q it points away from the element, along the line from the element through O.
  4. By symmetry the y-components of elements at +θ and −θ cancel. Only the x-component survives:

dEₓ = dE cosθ = (λ/4πε0 r) cosθ dθ

  1. Integrate from −60° to +60°:

Eₓ = (λ/4πε0 r) ∫ cosθ dθ = (λ/4πε0 r) [sinθ]₋₆₀°⁺⁶⁰° = (λ/4πε0 r)(2 sin 60°) = (λ/4πε0 r)√3

  1. Substitute λ = 3Q/(2πr):

E = (1/4πε0) · (3√3 / 2π) · Q / r²

  1. 3√3/(2π) = 5.196/6.283 = 0.827 ≈ 0.83

E = 0.83 Q/(4πε0 r²), directed along the axis of symmetry, away from the middle of the arc (for positive Q). Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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