The cube of insulating material shown in the figure has one

The cube of insulating material shown in the figure has one corner at the origin Fiach side of die cube length 0.080 m so the top face of the cube is parallel to the xz-plane and is at y 0.080 m It observed there is an electric field t - (3280N/Cm)y; that is in the + v direction and whose magnitude depends only on Use Gauss\'s law to calculate the net charge enclosed by the cube, (E0 = 8.85 x 10^-12 c2/N m^2)

Solution

in this case, the electric field is pointed in y-direction. the flux through

four surfaces of the cube is zero because electric field is parallel with

these four surfaces. the remaining two surfaces are perpendicular

to the electric field.

For the surface at y = 0, field is

            E = (3280 N/C m)(0) j

               = 0

hence, there is no flux through this surface.

For the surface at y = 0.080 m, the electric field is,

       E = (3280 N/C m)(0.080 m) j

          = 262.4 N/C j.

from Gauss\' Law, we have
                                EA = q/o

thus, the required net charge is,
        (262.4 N/C)(0.080 m)^2 = q/(8.85 *10^-12 C^2/N-m^2)
                                           q = 1.486 * 10^-11 C