Question

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

• A. $6.8\times {10}^5\text{C}$
• B. $6.8\times {10}^6\text{C}$
• C. $6.8\times {10}^4\text{C}$
• D. $6.8\times {10}^9\text{C}$

Answer 1

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

let us construct a gaussian sphere passing through surface of earth such that total charge of earth gets enclosed within it.

Given,
The electric field at earth's surface is, $E=150\text{V/m}$

Radius of the earth, $R=6400\text{km}=6400\times {10}^3\text{m}$

Applying gauss's law,

$\varphi =\overrightarrow{E}.d\overrightarrow{A}$ =$\frac{q_{in}}{{\epsilon }_o}$

$\Rightarrow EA=\frac{q_{in}}{{\epsilon }_o}$

$\Rightarrow E\times \left(4\pi R^2\right)=\frac{q_{in}}{{\epsilon }_o}$

Substituting the values,

$150\times 4\times \pi \times {(6400\times {10}^3)}^2=\frac{q_{in}}{{\epsilon }_o}$

$\Rightarrow q_{in}=150\times 4\times \pi \times {(6400\times {10}^3)}^2\times 8.85\times {10}^{-12}$

$\therefore q_{in}\approx 6.8\times {10}^5\text{C}$

Hence, total charge on the earth's surface is $6.8\times {10}^5\text{C}$.