Question

It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 260 m the field has magnitude 60.0 N/C. At an altitude of 150 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 110 m on edge, with horizontal faces at altitudes of 150 and 260 m. Neglect the curvature of Earth.

Answer 1

`1.176604*10^(-10) C`

Explanation:

This is flux problem (see diagram below)

Since electric field is vertical to the top and bottom of the cube. total flux will be sum of flux up and down i-e

Фtotal=Фup + Фdown = Ф 260 + Ф150

But flux is given by scalar product of Electric field and area

Ф=E.A= EA cos∅

where ∅ is the angle between two vector quantities E the electric field and A the area vector. So, total flux is

Фtotal =
`=E_(260) .A+E_(150) .A`

`=E_(260) *A cos \alpha +E_(150) *A cos \beta`

`=E_(260) * Acos(180)+E_(150) * Acos(0)`

`=-E_(260) .A +E_(150) .A`

`=A(E_(150) -E_(260) )`

But we know that according to Gausse's Law total flux is given by;

Total flux = q / ∈

==> q= ∈(Total flux)

`q = 8.84 *10^(-12) *110^(2) (260-150)`

`q=1.176604 *10^(-10) C`