Question

The electric field in a certain region of Earth’s atmo- sphere is directed vertically down. At an altitude of 300 m the field has magnitude 60.0 N/C; at an altitude of 200 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 100 m on edge, with horizontal faces at altitudes of 200 and 300 m

Answer 1

Answer:

Net charge contained in the cubeq= 3.536×10^-6C

Explanation:

Formular for total flux in a cube is given as:

Total flux= E300Acos(180) + E200Acos(0)

Where A is crossectional area

Total flux= A(E200-E300)

Total flux= q/Eo

q= Eo×total flux

q=(8.84×10^-12)×(100)^2×(100-60)

q= 3.536×10^-6C

Answer 2

Answer:

$q = 3.54*10^{-6}C$

Explanation:

Given that:

A = 100m

($E_l$) = 100

($E_v$) = 60

E₀(constant) = 8.85 * 10⁻¹²

The net amount of charge can be found using the expression:

=  $\frac{q}{E_0} = \theta$

=  $q= \theta E_0$   where θ =0; A(El-Ev)

=  $q= E_0A(E_l-E_v)$

=  $q=(8.85*10^{-12}*(100)^2*(100-60)$

=  $q=(8.85*10^{-12}*(100)^2*(40)$

=  $q=(8.85*10^{-12}*(400,000)$

=  $q = 3.54*10^{-6}C$