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.

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

newton's first law of motion

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.