It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 300 m the field has magnitude 60.0 N/C; at an altitude of 230 m, the magnitude is 100 N/C. Find the amount of charge contained in a cube 70 m on edge, with horizontal faces at altitudes of 230 and 300 m. Neglect the curvature of Earth.

q=1.7346×10⁻⁶C

Explanation:

Since the electric field is perpendicular to the bottom and top of the cube,the total flux is equals the flux over the top of surface plus the flex over the lower surface

Ф(total)=Ф₃₀₀+Ф₂₃₀

But the flux is given by  Ф=E.A=EACos(θ) where θ is the angle between Area vector and electric field

So

Ф(total)=E₃₀₀A Cos(180)+E₂₃₀ACos(0)

Ф(total)=A(E₃₀₀ - E₂₃₀)

The total flux is given by Gauss Law as:

Ф(total)=q/ε₀

q=ε₀Ф(total)

q=ε₀(A(E₃₀₀ - E₂₃₀))

Substitute the given values

q=(8.85×10⁻¹²){(70²)(100 - 60)}

q=1.7346×10⁻⁶C

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer