The Earth’s surface, on average, carries a net negative charge (while the clouds and lower atmosphere carry a net positive charge of nearly equal magnitude), giving rise to an electric field at its surface which is about 150 N/C in size.1 Assuming a spherical Earth, determine(a) the total charge on Earth’s surface responsible for this field and the corresponding surface charge density.(b) Given the 150 N/C field at Earth’s surface, how much lower is the field strength right below a cloud at a height of 5 km? [Careful for rounding errors here.]

(a) Q = -6.765 * 10⁵ C ; σ = 1.33 * 10⁻⁹ C/m²

(b) 0.23 N/C

Explanation:

(a) Electric field at the surface of a sphere is given as:

E = kQ/r²

Where

k = Coulombs constsnt

Q = charge

r = radius of sphere

To find charge Q, we make Q subject of the formula:

Q = (E * r²)/k

Hence, charge, Q, at the surface of the earth, having radius, r = 6.371 * 10⁵ and electric field, E = 150 N/C is:

Q = [150 * (6.371 * 10⁶)²] / (9 * 10⁹)

Q = 6.765 * 10⁵ C

Since we're told that the charge at the earth's surface is negative,

Q = -6.765 * 10⁵ C

Surface charge density, σ, given as:

σ = |Q|/A

Where

|Q| = magnitude of charge

A = surface area.

Surface area, A, of the earth is given as:

A = 4πr²

A = 4π * (6.371 * 10⁶)²

A = 510064471909788 m²

σ = 6.765 * 10⁵/510064471909788

σ = 1.33 * 10⁻⁹ C/m²

(b) At a height 5km from the earth's surface, the electric field will be:

E = kQ/(r + 5km)²

r + 5km = 6376km = 6.376 * 10⁶m

=> E = (9 * 10⁹ * 6.765 * 10⁵)/(6.376 * 10⁶)²

E = 149.77 N/C

The difference between the electric field at the surface of the earth and at a height of 5km is:

159 - 149.77 N/C = 0.23 N/C

yup its correct its: c) b b