Question

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150\,N/C$, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be: [Given, ${R_E} = 6.4 \times {10^6}m$
A) +660 kC
B) -660 kC
C) -680 kC
D) +680 kC

Answer 1

Hint: The charge enclosed inside the closed surface can be calculated from Gauss’s law. Assume the Earth as a point charge and calculate its electric field at a distance equal to the radius of the Earth.

Complete step by step solution:
We’ve been given that the average electric field present just outside the surface of the Earth is $150\,N/C$. Since we’re calculating the electric field outside the surface of the Earth, we can use Gauss’ law assuming a point charge containing the net charge carried by the Earth. Then for a point just above the surface of the Earth, the area of the Gaussian sphere will be equal to the area of the surface of the Earth. So, we can write
$\Rightarrow E.(4\pi {{R}^{2}}_{E})=\dfrac{q}{{{\epsilon }_{0}}}$
$\Rightarrow q={{\epsilon }_{0}}\times E.(4\pi {{R}^{2}}_{E})$
Substituting the value of ${R_E} = 6.4 \times {10^6}m$, and $E = 150\,N/C$, the value of
$\Rightarrow q = 8.85 \times {10^{ - 12}} \times 150 \times 4 \times \pi \times 6.4 \times {10^6} $
$\Rightarrow \,680\,kC$
However, since the electric field is directed inwards towards the Earth, the charge must be negative as the electric field lines are attracted to negative charges and we’ve been given that the electric field is directed inwards. So,
$q = - 680kC$ which corresponds to option (C).

Note:
Since the charge distribution on Earth’s surface is constant, we should assume it to be a point charge located at the center of the Earth which will simplify our calculations. While Gauss’s law will help us in determining the magnitude of the charge on the surface of the Earth, we must not forget to use the direction of the electric field to determine the polarity of the charge.