Nomal

The square surface shown in the figure measures 3.5 mm on each side. It is immersed in a uniform electric field with magnitude E = 2100 N/C and with field lines at an angle of 35° with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface. Normal 4

A cube of side L = 2.1 m lies in a region where the electric field is given by E = [-2(x^2/m^2+3.6) i - 2.0 k] N/C. We wish to find the net electric flux through the cube by first calculating the flux through each of the six faces. Here, x1=1.5m.   (a) What is the flux through the left face of the cube? (b) What is the flux through the right face of the cube? (c) What is the flux through the top face of the cube? (d) What is the flux through the bottom face of the cube? (e) What is the flux through the front face of the cube? (f) What is the flux through the back face of the cube? (g) What is the net flux through the cube?

An electric field given by E=3.01-9.6(y² +4.3) pierces the Gaussian cube of edge length 0.180 m and positioned as shown in the figure. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? (a) Number i -8e-6 Units N•m^2/C Gaussian surface -X