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?

Electric flux (Φ) through a surface is defined as the electric field (E) multiplied by the area (A) of the surface and the cosine of the angle (θ) between the electric field direction and the normal (perpendicular) to the surface: Φ = E * A * cos(θ).

For the first frame, which is placed perpendicular to the electric field (oriented along the y-axis) in the xz-plane, the angle θ between the electric field and the normal to the plane is 0 degrees, since the normal to the plane is along the y-axis. Therefore, cos(θ) = cos(0) = 1, and the electric flux through this frame is given as Φ0 = E * A * 1.

Now, in order to have the flux through the second frame be half of that of the first frame (Φ0/2), the product of the area and the cosine of the angle must be halved, assuming the electric field strength remains constant.

Since the area of the second frame is the same as the first (A), and the electric field (E) is uniform and unchanged, the only variable that can change is the angle (θ). We need to find the angle θ such that cos(θ) is one-half:

E * A * cos(θ) = Φ0/2 E * A * cos(θ) = (E * A * 1)/2 cos(θ) = 1/2

The angle with a cosine of 1/2 is 60 degrees (or π/3 radians).

Therefore, in order for the second frame to have half the electric flux, the plane must be oriented at an angle of 60 degrees with respect to the y-axis, which is the direction of the electric field. This means that the normal to the second frame makes a 60-degree angle with the y-axis.