Question

A uniform electric field ai + bj intersects a surface of area A. What is the flux through this area if the surface lies (a) in the yz plane? (b) in the xz plane? (c) in the xy plane?

Answer 1

Step-by-step explanation:

The electric flux through a surface is given by the dot product of the electric field and the area vector of the surface:

Φ = E · A

where Φ is the electric flux, E is the electric field, and A is the area vector of the surface.

(a) If the surface lies in the yz plane, its area vector is in the x direction. Therefore, the area vector can be written as A = Ax i, where Ax is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Ax i) = aAx

(b) If the surface lies in the xz plane, its area vector is in the y direction. Therefore, the area vector can be written as A = Ay j, where Ay is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Ay j) = bAy

(c) If the surface lies in the xy plane, its area vector is in the z direction. Therefore, the area vector can be written as A = Az k, where Az is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Az k) = 0

since the dot product of perpendicular vectors is zero.