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A current density of 7.00 10-13 A/m2 exists in the atmosphere at a location where the electric field is 185 V/m. Calculate the electrical conductivity of the Earth's atmosphere in this region.

Respuesta :

Answer:

electrical conductivity is given as

[tex]\sigma = 3.8 \times 10^{-15} [/tex]

Explanation:

For a current carrying conductor we will have relation between current density and electric field inside the conductor as

[tex]j = \sigma E[/tex]

here we have

current density = j

electric field = E

given that

[tex]j = 7.00 \times 10^{-13}[/tex]

[tex]E = 185 V/m[/tex]

now we will have

[tex]\sigma = \frac{j}{E}[/tex]

[tex]\sigma = \frac{7.00 \times 10^{-13}}{185}[/tex]

[tex]\sigma = 3.8 \times 10^{-15} [/tex]

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