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?

β = 30º

Explanation:

       Φ0 = E*A*cos 0º = E*A.

        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

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..