Question

What is the net charge of the Earth if the magnitude of its electric field near the terrestrial surface is 1.08 ✕ 102 N/C? Assume the Earth is a sphere of radius 6.40 ✕ 106 m.

Answer 1

To solve this problem we will apply the concepts related to the electric field based on the laws of Coulomb. Said electric field is equivalent to the product between the Coulomb constant and the rate of change of the charge and the squared distance. Mathematically this is,

$E = \frac{kq}{r^2}$

Here,

k = Coulomb's constant

q = Charge

r = Distance

Replacing we have that

$E = \frac{kq}{r^2}$

$1.08*10^2 = \frac{(9*10^{9})q}{(6.4*10^{6})^2}$

Solving for q,

$q = 491520 C$

Therefore the net charge of the Earth under the previous condition is 491520 C