During fair weather, an electric field of about 100 N/C points vertically downward into the earth's atmosphere. Assuming that this field arises from charge distributed in a spherically symmetric manner over the surface of the earth, determine the net charge of the earth and its atmosphere if the radius of the earth and its atmosphere is 6.37x10^6 m.
To determine the net charge of the Earth and its atmosphere, we need to use Gauss's Law, which relates the electric field to the total charge enclosed within a Gaussian surface. In this case, we can consider a Gaussian surface in the form of a sphere with a radius equal to the radius of the Earth and its atmosphere combined.
The electric field of 100 N/C pointing vertically downward into the Earth's atmosphere implies a net positive charge enclosed within the Gaussian surface. Since the electric field points in, we can assume a positive charge distribution on the surface of the Earth and its atmosphere.
The formula for the electric field inside a uniformly charged sphere is given by:
E = (k * Q) / r^2
Where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the net charge enclosed, and r is the radius of the Gaussian surface.
In this case, we'll assume uniform charge distribution on the surface of the Earth and its atmosphere, so the electric field inside is constant.
Given that the electric field is 100 N/C and the radius is 6.37 x 10^6 m, we can rearrange the formula to solve for the net charge Q:
Q = (E * r^2) / k
Plugging in the values:
Q = (100 * (6.37 x 10^6)^2) / (9 x 10^9)
Q ≈ 4.25 x 10^9 C
Therefore, the net charge of the Earth and its atmosphere is approximately 4.25 x 10^9 Coulombs.