Question

If the electric field intensity in fair weather atmosphere is $100\;{\text{V/m}}$ then the total charge on the earth’s surface is (radius of the Earth is $6400\;{\text{km}}$)
A) $4.55{\text{ }} \times {\text{ }}{10^7}\;{\text{C}}$
B) $4.55{\text{ }} \times {\text{ }}{10^8}\;{\text{C}}$
C) $4.55{\text{ }} \times {\text{ }}{10^5}\;{\text{C}}$
D) $4.55{\text{ }} \times {\text{ }}{10^6}\;{\text{C}}$

Answer 1

In this question, use the concept of the Coulomb’s law that is it state that the force exerted between the two charge is directly proportional to the product of the charges and inversely proportional to the square of the distance between the two charges.

Complete step by step solution:
In this question, we have given, the electric field intensity is $100\;{\text{V/m}}$, the radius of the Earth is $6400\;{\text{km}}$ we need to calculate the total charge on the Earth surface.
Let us consider the total charge on the Earth’s surface to be $Q$ coulombs. Let the radius be denoted as $r$ in meters, that is $r{\text{ }} = {\text{ }}6.4{\text{ }} \times {10^6}\;{\text{m}}$.
Now as per the definition of Electric Field Intensity, the Electric field Intensity at a point is the force that is experienced by a unit positive charge placed at that particular point. It is a vector quantity because it has both direction and magnitude. It is denoted by $E$.
$\Rightarrow E\; = \dfrac{F}{q}......\left( 1 \right)$
Now by coulomb’s law we have
$\Rightarrow F = \dfrac{{kQq}}{{{r^2}}}$
Now we substitute the expression of force in equation $\left( 1 \right)$ as,
$E = \dfrac{{\dfrac{{kQq}}{{{r^2}}}}}{q}$
On simplification we get,
$\Rightarrow E = \dfrac{{kQ}}{{{r^2}}}......\left( 2 \right)$
Here, the value of $E$ is given to be $100\;{\text{V/m}}$, radius of the earth $r{\text{ }} = 6.4{\text{ }} \times {10^6}\;{\text{m}}$, and the Coulomb’s constant .
Now we substitute the values in equation $\left( 2 \right)$ as,
$\Rightarrow 100 = \dfrac{{9 \times {{10}^9} \times Q}}{{{{\left( {6.4 \times {{10}^6}} \right)}^2}}}$
On simplification we get,
$\therefore Q = 4.55 \times {10^5}\;{\text{C}}$
Thus, the total charge on the earth’s surface is $4.55 \times {10^5}\;{\text{C}}$

Hence, the correct option is (C).

Note: Here the condition considered is a standard one, i.e. the normal atmospheric condition (fair weather condition) hence the value of coulomb’s constant is taken as the standard one. But if the same thing happens in a different dielectric medium then the value will be different. In that condition the value of permittivity will be different and hence the surface charge will also be different.