Question

The earth’s surface has a negative surface charge density of ${10^{ - 9}}\,C{m^{ - 2}}$. The potential difference of $400\,kv$ between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only $1800\,A$ over the entire globe. If there were no mechanism of sustaining an atmospheric electric field, how much time (roughly) would be required to neutralize the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth $ = 6.37 \times {10^6}\,m$).

Answer 1

Hint: Charge density is defined as the amount of the charge that is present in the unit length of the conductor. Total charge can be found by the product of the charge density and the area of the conductor. Current is obtained by dividing the charge by the time taken.

Formula used:
(1) Area of sphere is given by
$A = 4\pi {r^2}$
Where $A$ is the area of the sphere and $r$ is the radius of the sphere.
(2) The charge formula is given by
$q = \sigma A$
Where $\sigma $ is the charge density and $q$ is the charge.
(3) The current formula is
$I = \dfrac{q}{t}$
Where $I$ is the current flowing through the surface, $t$ is the time taken for a flow.

Complete step by step answer:
Given: Surface charge density, $\sigma = {10^{ - 9}}\,C{m^{ - 2}}$
The current over the full globe, $I = 1800\,A$
Radius of the earth, $r = 6.37 \times {10^6}\,m$
Since the earth is the sphere, the surface area of the earth is calculated by the formula of the area of the sphere,
$A = 4\pi {r^2}$
Substituting the values in it,
$A = 4 \times \pi \times {\left( {6.37 \times {{10}^6}} \right)^2}$
$A = 5.09 \times {10^{14}}\;{m^2}$
Let us take the formula of the charge,
$q = \sigma A$
Substituting the values in it,
$q = 5.09 \times {10^5}\,C$
Let us consider the formula of the current,
$I = \dfrac{q}{t}$
$t = \dfrac{{5.09 \times {{10}^5}}}{{1800}} = 282\,s$

Hence, The time required to neutralize the earth’s surface is $282\,s$.

Note: The earth surface contains the electrical charges in it which is gained from the thunder and the lightning. Hence in practical analysis the neutralizing of all the electrical charges on the earth is not possible, however it can be again and again formed.