The electric field in a certain region of Earth's atmosphere 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.

Question

Alfred Butler

Physics

9 March, 21:35

The electric field in a certain region of Earth's atmosphere 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.

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Answers (1)

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Carrillo · Answer 1

q = 3.356*10^-6C

Explanation:

In a cube of air, the electric field is perpendicular to the top and bottom of the cube, the total reflux = reflux200 + reflux300

But reflux is given by:

Reflux = E*A = E*A cos theta.

Where theta is the angle between the area vector and the electric field.

Total reflux = E300*A cos180 + E200*A cos0

Total reflux = - E300A + E200A

Total reflux = A (E300-E200)

Total reflux is given by Guass's law = Total reflux = q/Eo

q=Eo*total reflux

q = EoA (E200-E300) q = (8.84*10^-12) (100) ^2 * (100-60)

q=3.356*10^-6C