The electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 241 m the field has magnitude 46.8 N/C; at an altitude of 93.9 m, the magnitude is 114 N/C. Find the net amount of charge contained in a cube 147.1 m on edge, with horizontal faces at altitudes of 93.9 m and 241 m.

Explanation:

We shall apply Gauss's Law to find the solution .

flux entering the cube = E X A

= 46.8 X 147.1 X 147.1 Weber

flux going out of cube

= 114 x  147.1 X 147.1  Weber

Net flux going out

= ( 114 x  147.1 X 147.1  - 46.8 X 147.1 X 147.1 )

= 1454101.15 weber .

according to Gauss's law

q / ε₀ = 1.4541 x 10⁶ , q is required charge inside the cube.

q = 8.85 x 10⁻¹² x 1.4541 x 10⁶

= 12.868 x 10⁻⁶ C

= 12.868 μC.

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!

the right answer is 0.3 meters per second