Question

what do you think the electrical flux through each of the following cylindrical surfaces (oriented outward) placed in a constant vector field will be? positive, negative, or zero? why? be sure you can explain your answer. a manipulable 3d graph. a manipulable 3d graph. a. choose b. choose

Answer 1

Electrical flux through a cylindrical surface in a constant vector field is positive or negative on the end caps depending on field orientation, and zero through the sides. The net electric flux through a closed cylindrical surface in such a field is zero unless it encloses a charge.

Step-by-step explanation:

When considering the electrical flux through a cylindrical surface in a constant vector field, the outcome depends on the orientation of the surface relative to the direction of the electric field. If the cylindrical surface's flat end caps are perpendicular to the field lines, the flux will be positive or negative depending on the direction of the field relative to the outward normal of the surface; positive if they are in the same direction, negative if they are in the opposite direction. The flux through the cylindrical side surface is zero because the electric field is perpendicular to the surface, and therefore no field lines pass through it. According to Gauss's Law, the net electric flux through any closed surface is proportional to the enclosed charge. In a constant vector field with no enclosed charge, the net flux through a surface would be zero.

Answer 2

The electrical flux through the end caps of a cylindrical surface in a constant vector field will be positive or negative depending on the direction of the electric field relative to the normal of the surface. The sides of the cylinder have zero flux because they are parallel to the field lines. Overall, for a cylinder in a uniform field with no enclosed charge, the net flux is zero.

Step-by-step explanation:

When considering electrical flux through a cylindrical surface in a constant vector field, we should first understand that the flux is determined by how the electric field lines intersect the surface. If the cylindrical surface's end caps are perpendicular to the electric field, the flux through these end caps would be maximal because the field lines and surface normals are aligned. However, for the cylindrical part of the surface (the sides), the flux would be zero since the surface is parallel to the field lines.

In terms of sign, if the electric field lines are exiting the cylinder (that is, the direction of the field is the same as the outward normal to the end cap), the flux would be positive. If the field lines are entering the cylinder, the flux would be negative. This is consistent with Gauss's Law, which states that the total electric flux out of a closed surface is proportional to the enclosed charge. For a cylinder in a uniform field with no charge enclosed, however, the net flux across both end caps (one positive and one negative) would cancel out, resulting in a net flux of zero.

To illustrate with a manipulable 3D graph, imagine the cylinder oriented with one end facing towards the field and the other away. The flux through each end would be equal and opposite, and when summed, the total would be zero unless there is a charge enclosed within the cylinder.