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Field of charged rod along its axis

Short answer

Using linear density λ = Q/l, integrate dE = (1/4πε₀) λ dx/x² from x=a to a+l, giving E = Q/(4πε₀ a(a+l)), directed along the rod away from it.

Calculate the electric field due to a uniformly charged rod of length l at a point along its long axis at a distance 'a' from its nearest end.

Rod of length l, total charge Q, uniform linear density λ = Q/l. Find the field at point P on the line of the rod, a distance a from the nearest end.

  |==========================|        . P
  0                          l   <-a->
  x axis along rod ->
  dq at distance x from P

Measure distance x from P along the rod, so the rod extends from x = a to x = a + l.

Steps

  1. Take an element of length dx at distance x from P, carrying dq = λ dx.
  2. Its field at P is along the rod (all elements are collinear with P, so contributions add directly):

dE = (1/4πε0) λ dx / x²

  1. Integrate over the rod:

E = (λ/4πε0) ∫ₐ^(a+l) dx/x² = (λ/4πε0) [−1/x]ₐ^(a+l) = (λ/4πε0) [1/a − 1/(a+l)]

  1. Simplify: 1/a − 1/(a + l) = l / (a(a + l))

E = (λ l) / (4πε0 a (a + l))

  1. Since λ l = Q:

E = Q / (4πε0 a (a + l)), directed along the rod, away from it for positive Q.

Check: for a ≫ l, E → Q/(4πε0 a²), the field of a point charge.

✓ Answer checked by Claude Opus 5.5 (high effort)

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