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Field of long charged rod, equatorial

Short answer

General method: divide into dq = λ dl, treat as point charges, integrate E = ∫dE and V = ∫dV using symmetry. For a long rod, equatorial field: E = λ/(2πε₀ y) perpendicular to the rod, directed away for λ>0.

Give a general method to calculate electric field and potential due to continuous charge distribution. Using your method, calculate electric field at an equatorial distance y due to a long charged rod having linear charge density λ.

General method for a continuous distribution

  1. Divide the distribution into small elements dq (dq = λ dl for a line, σ dA for a surface, ρ dV for a volume).
  2. Treat each dq as a point charge. Its field is dE = (1/4πε0) dq/r² along the line from dq to the point; its potential is dV = (1/4πε0) dq/r.
  3. Resolve dE into components and use symmetry to cancel those that sum to zero.
  4. Integrate: E = ∫dE (vector sum); V = ∫dV (scalar sum, easier). Check by E = −∇V.

Long charged rod, field at equatorial distance y

Rod of length 2L (very long), uniform λ, point P at perpendicular distance y from the middle.

                P
               /|\
            r / |y \ r
             /  |    \
  ----------o---O-----o----------   rod
          -x    0    +x
  1. Element dq = λ dx at x from the centre. r = √(x² + y²).
  2. dE = (1/4πε0) λ dx / r². The x-components of elements at +x and −x cancel. Component along y: dE_y = dE cosθ, with cosθ = y/r.
  3. Put x = y tanθ, dx = y sec²θ dθ, r = y secθ:

dE_y = (λ/4πε0) (y sec²θ dθ / y² sec²θ) cosθ = (λ/4πε0 y) cosθ dθ

  1. Integrate from −θ₀ to +θ₀ (angles subtended by each end):

E = (λ/4πε0 y) · 2 sinθ₀ = λ L / (2πε0 y √(L² + y²))

  1. For a long rod (L ≫ y), θ₀ → 90°, sinθ₀ → 1:

E = λ / (2πε0 y), perpendicular to the rod, directed away for λ > 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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