Question

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.

Answer 1

Step-by-step explanation:

Electric field at height 241 m , E = 46.8 N/C

Electric field at height 93.1 m, E' = 114 N/C

length of edge, a = 147.1 m

The total flux

Фnet = Ф + Ф'

Фnet = E A Cos 180 + E' A Cos 0

Фnet = A ( E' - E)

According to the Gauss theorem

Фnet = enclosed charge / ∈o

∈o x A (E' - E) = q

q = 8.854 x 10^-12 x 147.1 x 147.1 x (114 - 46.8)

q = 1.29 x 10^-5 C

Answer 2

Step-by-step 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.