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 .

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

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6