Question

The electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 298 m the field has magnitude 48.4 N/C; at an altitude of 166 m, the magnitude is 109 N/C. Find the net amount of charge contained in a cube 132 m on edge, with horizontal faces at altitudes of 166 m and 298 m.

Answer 1

Answer:

$F_{total}= E_{298m} A cos (180) + E_{166m} A cos (0)$

$F_{total}= A (E_{166 m}- E_{298 m})$

Using the Gauss's Law defined as:

$F_{total}= \frac{q}{\epsilon_o}$

We have that:

$q = \epsilonn_o F_{total}$

And replacing into the formula for the total flux we got:

$q = \epsilon_o A (E_{166 m}- E_{298 m})$

And replacing we got:

$q = 8.84x10^{-12} (132m*132 m) (109 -48.4) N/C= 9.33x10^{-6}C$

Explanation:

For this case we need to find the total charge inside a cube of air with sides of length 132 m, the top altitude would be 298 m and the lower altitude would be 166 m. We know that :

$E_{298 m}= 48.4 N/C, E_{166 m}= 109N/C$

The figure attached illustrate the problem.

For this case the electric field would be perpendicular to the bottom and the top of the cube. So the total flux on this case would be the addition of the two fluxes like this:

$F_{total} = F_{166 m}+ F_{298 m}$

By definition the flux is given by this formula:

$F = E A = E A cos (\theta)$

Where $\theta$ represent the angle between the area vector and the electric field.

Replacing this we got:

$F_{total}= E_{298m} A cos (180) + E_{166m} A cos (0)$

$F_{total}= A (E_{166 m}- E_{298 m})$

Using the Gauss's Law defined as:

$F_{total}= \frac{q}{\epsilon_o}$

We have that:

$q = \epsilon_o F_{total}$

And replacing into the formula for the total flux we got:

$q = \epsilon_o A (E_{166 m}- E_{298 m})$

And replacing we got:

$q = 8.84x10^{-12} (132m*132 m) (109 -48.4) N/C= 9.33x10^{-6}C$