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

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

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