Question

A flat circle of radius 0.18 is placed in a uniform electric field of magnitude 5.13×10² N/C. What is the electric flux through the circle when its face is 45° to the field lines?

Answer 1

Consider the following figure: Flux is total number of field lines passing through an area perpendicular to the field lines. So, we take the component of fields that are perpendicular to the given area. The flux through the circle depends on the angle between normal to the circular disk and the direction of electric field. If the angle is given to be theta, and the electric field is uniform, then the flux is given by: pbhi = EA cos theta where A is the area through which the field lines are passing. Here, A will be the area of the circle, given by, pi r^2. Substitute the values to get: phi = E pi r^2 cos theta = (5.13 times 10^2) (pi 0.18^2) cos 45 = 36.9 Nm^2/C