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15 November, 16:27

A current density of 8.50 10-13 A/m2 exists in the atmosphere at a location where the electric field is 143 V/m. Calculate the electrical conductivity of the Earth's atmosphere in this region.

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  1. 15 November, 16:38
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    The electrical conductivity is 5.94*10^-15 ohm^-1 m^-1

    Explanation:

    Electrical conductivity = current density : electric field

    Current density = 8.5*10^-13 A/m^2

    Electric field = 143 V/m

    Electrical conductivity = 8.5*10^-13 A/m^2 : 143 V/m = 5.94*10^-15 ohm^-1 m^-1
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