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

Answer:

$1.176604*10^{-10} C$

Step-by-step 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)

$q = 8.84 *10^{-12} *110^{2} (260-150)$

$q=1.176604 *10^{-10} C$