91Ó°ÊÓ

1. At an altitude of 300 m, the field has magnitude 60.0 N/C.
2. At an altitude of 200 m, the magnitude is 100 N/C .
3. Cube 100 m on edge, with horizontal faces at altitudes of 200 and 300 m.

Using the concept of the Gauss flux theorem, we can get the net flux passing through the surface by substituting the net electric field through the surface.

Formula:

The electric flux passing through the surface,

$\mathrm{Ï•}=\stackrel{→}{dE}.\stackrel{→}{A}=\frac{q}{{\mathrm{Î\mu }}_0}$ (1)

Let A be the area of one face of the cube, ${\mathrm{E}}_{\mathrm{u}}$be the magnitude of the electric field at the upper face, and ${\mathrm{E}}_1$be the magnitude of the field at the lower face. Since the field is downward, the flux through the upper face is negative and the flux through the lower face is positive. The flux through the other faces is zero (because their area vectors are parallel to the field), so the total flux through the cube surface is given using equation (1) as:

$\mathrm{Ï•}=\mathrm{A}({\mathrm{E}}_1-{\mathrm{E}}_{\mathrm{u}})$

Using the equation (1), we can get the enclosed charge by the surface of the cube as:

$$\begin{gathered} \mathrm{q}={\mathrm{Î\mu }}_0\mathrm{Ï•} \\ ={\mathrm{Î\mu }}_0\mathrm{A}({\mathrm{E}}_{\mathrm{i}}-{\mathrm{E}}_{\mathrm{u}}) \\ =(8.85×{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2){(*100\mathrm{m})}^2(100\mathrm{N}/\mathrm{C}-60.0\mathrm{N}/\mathrm{C}) \\ =3.54×{10}^{-6}\mathrm{C}\left(\frac{1\mathrm{μ°ä}}{{10}^{-6}\mathrm{C}}\right) \\ =3.54\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the net charge is $3.54\mathrm{μ°ä}$