A current density of 2.5 A m⁻² is found to exist in a conductor when an electric field of 5×10⁻⁸ V m⁻¹ is applied across it. The resistivity of the conductor is .........

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(2) 2×10⁻⁸ Ω m. Given: j = 2.5 A m⁻², E = 5×10⁻⁸ V m⁻¹. Idea: Ohm's law in its microscopic form is E = ρ j, so ρ = E/j. ρ = (5×10⁻⁸)/(2.5) = 2×10⁻⁸ Ω m. Units: V m⁻¹ divided by A m⁻² is V A⁻¹ m, that is Ω m.

A current density of 2.5 A m⁻² is found to exist in a conductor when an electric field of 5×10⁻⁸ V m⁻¹ is applied across it. The resistivity of the conductor is .........

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