Answer:
(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
