During fair weather conditions atmosphere of the Earth can be considered as a good conductor with resistivity $ρ$ depending on the height h from the surface of the Earth as $\frac{1}{ρ} = a + b h^{2},$ where a and b are constants. It is also known that near the surface of the earth electric field is $E_{0}$ directed towards the center of the earth which causes a radial current in the atmosphere. Assume $E_{0}$ is constant in magnitude and h<<R.

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The electrons will be travelling in radially outward direction.

The charge density will be negative at any general height.

The charge density at height h is $\frac{2 ε_{0} E_{0}}{R + h}$ in magnitude.

The current density J decreases with increase in height.

answer is B, C, D.

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Detailed Solution

$J = σ E = \frac{1}{ρ} E$
$J = (a + b h^{2}) E_{0}$
Hence option (A) is incorrect. Current density increases with increase in height.
As field is directed downward electrons will flow upward (radially outward).
Now, to calculate charge density $(ρ)$ we use gauss law (Assume a thin spherical shell as gaussian surface)
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