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Asked in GUJCET 2022 · Charged cylinder and line charge
Given: E₁ = 9×10⁴ N C⁻¹ at r₁ = 2 cm; find E₂ at r₂ = 3 cm.
Idea: the field of an infinite line charge is E = λ/(2πε₀ r), so E ∝ 1/r.
(E₂)/(E₁) = (r₁)/(r₂) = 2/3
E₂ = 9×10⁴×2/3 = 6×10⁴ N C⁻¹
Only the ratio of the distances enters, so the centimetres need not be converted. (The point-charge law 1/(r²) would give 4×10⁴, which is not offered: a line charge's field falls off more slowly.)
So the field at 3 cm is 6×10⁴ N C⁻¹.
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