Question

The figure shows a closed Gaussian surface in the shape of a cube of edge length 1.8 m, with one corner at x₁ = 4.4 m,y₁ = 4.3 m. The cube lies in a region where the electric field vector is given by E→ = - 2.4 î - 3.2 y² ĵ + 3.4 k̂ N/C, with y in meters. What is the net charge contained by the cube?

Answer 1

Given

cube of edge lenght l = 1.8 m

with one corner at x1 = 4.4 m,y1 = 4.3 m.

The cube lies in a region where the electric field vector is given by

E = - 2.4 i - 3.2 y2 j + 3.4 k N/C, with y in meters

by definition of electric flux phi = E*A cos theta where theta is the angle between E and normal to the surface area A

and by Gauss law the net electric flux through the closed surface area equal to (1/epsilon0)times the charge enclosed by the closed surface area

   so E*A cos theta = q_in / (1/epsilon0)

first we should calculate the net electric field through the cube is

E = E1+E2

E = - 2.4 i - 3.2 y2 j + 3.4 k N/C means the field lines are along y axis

so the field along +y axis is E1 = -3.2*4.3^2 = -59.168 N/C

now the flux is phi_1 = E1*A1 cos theta1 = -59.168*1.8^2*cos 180 = -191.70432 N.m^2/C

and phi2 along -y direction (left) is

   phi2 = -3.2*1.8^2*1.8^2 cos0 = +34 N*m^2/C

net flux is phi = -191.70+34 N/C = -157.7 N*m^2/C

now the charge is Q_in = -157.7*8.854*10^-12 C = (-1.3962758)*10^-9 C