Question

If the atmospheric electric field is approximately $150\text{V/m}$ and the radius of the earth is $6400\text{km}$, then the total charge of the earth is:-

• A. $6.8\times {10}^5\text{C}$
• B. $6.8\times {10}^6\text{C}$
• C. $6.8\times {10}^8\text{C}$
• D. $6.8\times {10}^9\text{C}$

Answer 1

The correct option is A $6.8\times {10}^5\text{C}$

Given that,
Atmospheric electric field, $\overrightarrow{E}=150\text{V}$
Radius of earth, $R=6400\text{km}=6400\times {10}^3\text{m}$

Let us draw a spherical Gaussian surface of radius $R$ passing through the atmosphere, such that the total charge of the earth gets enclosed within it.

Let $q_{in}$ be the total charge enclosed and $dA$ be an elemental area of the Gaussian surface.


Applying Gauss's law,
$\oint \overrightarrow{E}.\overrightarrow{dA}=\frac{q_{in}}{{ϵ}_o}$
Since the atmospheric electric field will be perpendicular to the Gaussian surface,
$\Rightarrow EA=\frac{q_{in}}{{ϵ}_o}$
$\Rightarrow E\times 4\pi R^2=\frac{q_{in}}{{ϵ}_o}$
$\Rightarrow 150\times 4\pi \times (6400\times {10}^3{)}^2=\frac{q_{in}}{{ϵ}_o}$
$\Rightarrow q_{in}=150\times 4\pi \times (6400\times {10}^3{)}^2\times 8.8\times {10}^{-12}$
$\therefore q_{in}=6.79\times {10}^5\text{C}\approx 6.8\times {10}^5\text{C}$

Hence, option (a) is correct.

Why this question? Note: The approximate charge enclosed in earth suggests that the earth is a very huge reservair of charge