## A current density of 7.3 × 10−13 A/m2 exists in the atmosphere where the electric field (due to charged thunderclouds in the vicinity) is 48

Question

A current density of 7.3 × 10−13 A/m2 exists in the atmosphere where the electric field (due to charged thunderclouds in the vicinity) is 48 V/m . Calculate the electrical conductivity of the Earth’s atmosphere in this region. Answer in units of Ω−1 · m −1 .

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3 weeks 2021-08-25T19:52:13+00:00 1 Answers 1 views 0

σ = 1.52 10⁻¹²   1 /Ω m

Explanation:

The current density is linear in the atmosphere, so it complies with Ohm’s law

J = σ E

Where J is the current density, sigma is the electrical inductance and E is the electric field

σ = J / E

σ = 7.3 10⁻¹³ / 48

σ = 1.52 10⁻¹² A / V 1 / m

σ = 1.52 10⁻¹²   1 /Ω m