An insulating cylindrical rod of diameter d and length ℓ, with ℓ ≫ d, has uniform surface charge density. The electric field just outside its curved surface at M is E₀. At a distant point P, where r ≫ ℓ, the field is E = E₀dℓ/(nr²). Find n.

Question

G1 Physics · Accepted Answer

For the long charged cylinder, E₀ = σ/ε₀, so σ = ε₀E₀. Its curved-surface charge is Q = σπdℓ = ε₀E₀πdℓ. At r ≫ ℓ it behaves as a point charge: E = Q/(4πε₀r²) = E₀dℓ/(4r²). Therefore n = 4.