Question

6. The Earth’s electric field is directed vertically downward; at the ground level its strength is E = 130 N/C, and at an altitude of 100 m E = 70 N/C. Find the charge in a cube 100 m on edge located between the ground and a 100 m altitude.

Answer 1

Answer:

5.31x10⁻⁶ C

Explanation:

The cube is located 100 m altitude from the ground, so the superior face is at 100m and has E = 70 N/C, and the inferior face is at the ground with E = 130 N/C.

The electric field is perpendicular to the bottom and the top of the cube, so the total flux is the flux at the superior face plus the flux at the inferior face:

Фtotal = Ф100m + Фground

Where Ф = E*A*cos(α). α is the angle between the area vector and the field (180° at the topo and 0° at the bottom):

Фtotal = E100*A*cos(180°) + Eground*A*cos(0°)

Фtotal = 70A*(-1) + 130*A*1

Фtotal = 60A

By Gauss' Law, the flux is:

Фtotal = q/ε, where q is the charge, and ε is the permittivity constant in vacuum = 8.854x10⁻¹² C²/N.m²

A = 100mx100m = 10000 m²

q = 60*10000*8.854x10⁻¹²

q = 5.31x10⁻⁶ C