Problem 33.

A cube of side 1 cm is placed within an electric field of  100 N/C. What is the flux through the (a) top face (b) bottom face (c) right face (d) left face of the cube. What is the net flux through the cube. The electric field is perpendicular to the left face of the cube.

Solution

To characterize the orientation of the faces of the cube we introduce the normal to the surface (faces) of the cube as shown in the figure below. Then the flux through any surface can be calculated as

where is the magnitude of the electric field, is the area of the surface, and is the angle between the direction of the electric field and the direction of the normal to the surface.

(a) Top face: The normal is perpendicular to the electric field. Then and . Then the flux is 0.

(b) Bottom face: The normal is perpendicular to the electric field. Then and . Then the flux is 0.

(c) Right face: The normal to the right face has the same direction as the direction of the electric field. Then and . The area of the face of the cube is (we need to use the correct units for the side of the cube: 1 cm = 0.01 m):

Then the flux is

(d) Left face: The direction of the normal to the left face is opposite to the direction of the electric field. Then and . The area of the face of the cube is (we need to use the correct units for the side of the cube: 1 cm = 0.01 m):

Then the flux is

The net flux through the cube is proportional to the net charge inside the cube (Gauss's law). Since the charge inside the cube is zero then the net flux is zero.