The electric field in a certain region of earth's atmosphere is directed vertically down. at an altitude of 439 m the field has magnitude 50.7 n/c; at an altitude of 361 m, the magnitude is 84.3 n/c. find the net amount of charge contained in a cube 78 m on edge, with horizontal faces at altitudes of 361 m and 439 m.

1.8* 10^-6 C

Explanation: Considering the flux for the electric in the two surface corresponding to the cube.

For the top surface, the flux is = 50.7 N/C* 78*78 m^2= 50.7*6084=

-3.084*10^5 C*m^2 ( the electric field and the normal to the surface have opossite directions)

For the bottom surface, the flux is 84.3 N/C * 78*78 m^2= 84.3*6084=

5.128 *10^5 C m^2

By the Gauss Law, the net flux of the electric field is the total charge closed by the gaussian surface.

So the net flux is 2.044 * 10^5 C*m^2

The net flux= Q/ε0 then

the total charge is  2.044 * 10^5 C*m^2 * 8.85 * 10^-12= 1.8* 10^-6 C

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

connect the batteries in series and hope