A circular arc of radius R has fixed linear charge density λ and subtends angle θ at its centre. Using E = (2kλ/R) sin(θ/2),what happens to the field at the centre as θ → 0?
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A circular arc of radius R has fixed linear charge density λ and subtends angle θ at its centre. Using E = (2kλ/R) sin(θ/2),what happens to the field at the centre as θ → 0?

G1 Physics · Accepted Answer

For θ ≪ 1, sin(θ/2) ≈ θ/2. Hence, E ≈ (2kλ/R)(θ/2) = kλθ/R → 0. This is physically justified because λ is fixed, so the total charge of the shrinking arc is Q = λRθ → 0.