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 .

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

Answers ( )

    0
    2021-08-25T19:53:23+00:00

    Answer:

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )