Question

A region of space contains a uniform electric field oriented along the y-axis. A frame of surface area A is placed perpendicular to the y-axis in the xz-plane. The magnitude of the electric flux through this frame is Φ0. A second frame is placed in the same electric field in such a way that the magnitude of the electric flux through it is Φ0/2. How is the plane of second frame oriented with respect to the plane of the first one?

Answer 1

β = 30º

Step-by-step explanation:

• By definition, the vector flux across a surface, can be found integrating the dot product of the vector field (electric field E in this case) and the differential surface dA, across the entire surface.
• If the surface is placed perpendicular to the electric field, and this field is uniform, the total flux across the surface can be expressed as follows:

Φ0 = E*A*cos 0º = E*A.

• If the magnitude of the electric flux, is reduced to half of its original value, we can write the following equality:

Φ0/2 = E*A*cos θ⇒ = Φ0 * cos θ (1)

where θ, is the angle between the electric field E and the vector

perpendicular to the plane traversed by E.

Rearranging terms in (1) we can solve for θ, as follows:

⇒ cos θ = 1/2 ⇒ θ = arc cos (1/2) = 60º

As this is the angle between the electric field, and the surface vector,

which is by definition, perpendicular to the plane, the angle between

the electric field and the plane can be found as follows:

β = 90º - θ = 90º - 60º = 30º