This question was asked inGPSC AE (Electrical) Class-2 R&B Department Paper 2025 – Series MQuestion 42 of 179 published · Gujarat Public Service CommissionOpen paper
Q151EasyElectrostatics

Calculate the conduction current density when the resistivity of a material with an electric field of 5 units is 4.5 units.

Answer & step-by-step solution
Correct answer: C. 5/4.5
Quick explanation

Conduction current density is related to electric field by the point form of Ohm’s law: J = σE = E/ρ. With E = 5 and resistivity ρ = 4.5, the current density is 5/4.5 ≈ 1.11 in the corresponding current-density units.

σ = 1/ρ J = σE = E/ρ J = 5/4.5 ≈ 1.11

Concept & Reasoning

Resistivity tells us how strongly a material opposes conduction. Conductivity is its reciprocal, σ = 1/ρ. Therefore a finite electric field produces a conduction current density J = σE. The correct numerical value follows directly from dividing the field intensity by resistivity; multiplying E by ρ would reverse the physical relationship.

How to Solve It in the Exam

Remember: resistivity in denominator, conductivity in numerator.

Important Exam Point

Do a dimensional/physical sense check: larger resistivity must reduce J for the same E.

Quick Trick

Remember: resistivity in denominator, conductivity in numerator.

Common Mistake

Using J = ρE instead of J = E/ρ.

Solution diagram – Calculate the conduction current density when the resistivity of a material…
Exam tipDo a dimensional/physical sense check: larger resistivity must reduce J for the same E.
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