A cube is located with one corner at the origin of an x, y, z, coordinate system. One of the cube's faces lies in the x, y plane, another in the y, z plane, and another in the x, z plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are 0.20 m long. A uniform electric field is parallel to the x, y plane and points in the direction of the +y axis. The magnitude of the field is 1500 N/C. Find the electric flux through each of the six faces of the cube.
a. bottom face (N·m2/C)
b. top face( N·m2/C)
c. each of the other four faces( N·m2/C)
flux= integral E.dA or E Area perpendicular to the E field. Each side has an area of .04m^2. So if the E field is in y direction, you deal with the xz planes.
Sketch the cube, if the E is going inward, the flux is negative, outward positive.
To find the electric flux through each face of the cube, we need to use Gauss's law which states that the electric flux through a closed surface is equal to the net charge enclosed divided by the permittivity of free space (ε0).
First, let's analyze the symmetry of the cube. Since the electric field is parallel to the x, y plane and points in the +y direction, the electric field lines are perpendicular to the cube's x, y face but parallel to the other faces. This means that only the x, y face of the cube will contribute to the electric flux.
a. Bottom face:
The electric field lines are completely perpendicular to the bottom face, so the electric flux passing through it will be zero. This is because the dot product between the electric field vector (E) and the normal vector (A) of the face will be zero.
Flux through bottom face = 0 N·m²/C
b. Top face:
The electric field lines are also perpendicular to the top face, and since the magnitude of the electric field is uniform, the electric flux will be the same as the bottom face.
Flux through top face = 0 N·m²/C
c. Each of the other four faces:
The electric field lines are parallel to these faces, so the electric flux will be non-zero. These faces have the same area and are perpendicular to the y-axis.
To find the electric flux through the other four faces, we need to calculate the net charge enclosed by each face. Since the electric field is constant, the flux through each face is the same.
Net charge = Total charge inside the cube = Total charge per unit volume * Volume of the cube
Since the cube is symmetric, it is enough to focus on one face. The area of each face of the cube is (0.2 m)² = 0.04 m².
The volume of the cube = (0.2 m)³ = 0.008 m³.
The total charge per unit volume has not been given in the problem, so you need that information to calculate the net charge enclosed by each face.
Once you have the net charge, you can calculate the electric flux through each face using Gauss's law:
Flux through each of the other four faces = Net charge enclosed / ε0
However, please note that without the charge density information or the ability to assume a charge distribution, we are unable to provide an exact numerical value for the flux through each of the other four faces.
To find the electric flux through each of the six faces of the cube, we can use Gauss's law. Gauss's law states that the electric flux passing through a closed surface is equal to the electric charge enclosed divided by the permittivity of free space.
In this case, since the cube is located with one corner at the origin, the electric field passing through the cube will be uniform and perpendicular to each of the faces. This means that the electric flux passing through each face will be equal.
The formula for electric flux is given by:
Electric flux = Electric field * Area of the surface * Cosine of the angle between the electric field and the normal vector of the surface
a. The bottom face of the cube lies in the x, y plane. The normal vector of the bottom face is in the -z direction. Since the electric field is parallel to the x, y plane, the angle between the electric field and the normal vector is 90 degrees. Therefore, the cosine of the angle is 0.
Electric flux through the bottom face = Electric field * Area of the bottom face * Cosine(90 degrees)
= 1500 N/C * (0.20 m * 0.20 m) * 0
= 0 N·m²/C
b. The top face of the cube lies in the x, y plane. The normal vector of the top face is in the +z direction. Since the electric field is parallel to the x, y plane, the angle between the electric field and the normal vector is also 90 degrees. Therefore, the cosine of the angle is 0.
Electric flux through the top face = Electric field * Area of the top face * Cosine(90 degrees)
= 1500 N/C * (0.20 m * 0.20 m) * 0
= 0 N·m²/C
c. Each of the other four faces of the cube lies in either the y, z plane or the x, z plane. The normal vectors of these faces are either in the -x, +x, -y, or +y directions. Since the electric field is parallel to the x, y plane, the angles between the electric field and the normal vectors of these faces are 90 degrees. Therefore, the cosine of the angles is 0.
Electric flux through each of the other four faces = Electric field * Area of each face * Cosine(90 degrees)
= 1500 N/C * (0.20 m * 0.20 m) * 0
= 0 N·m²/C
Therefore, the electric flux through each of the six faces of the cube is 0 N·m²/C.