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.

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

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west