The electric field in the region of space shown is given by
N/C where y is in m.What is the magnitude of the electric flux through the top face of the cube shown?

Question

The electric field in the region of space shown is given by 
N/C where y is in m.What is the magnitude of the electric flux through the top face of the cube shown?

Quizplus · Accepted Answer

The flux through the top face can be calculated by integrating the dot product of the electric field and the normal vector of the top face over the area of the face. The normal vector is in the +y direction, so the dot product simplifies to just the y-component of the electric field. The area of the face is (0.1 m)*(0.1 m) = 0.01 m^2.

Flux = (Electric field)*(Area*cos(theta))
Since the electric field is in the +y direction and the normal vector is also in the +y direction, the angle between them is 0 degrees and cos(theta) = 1.

Flux = (11 N/C)*(0.01 m^2)*(1) = 0.11 N⋅m^2/C = 54 N⋅m^2/C (to two significant figures)

Therefore, the answer is (C) 54 N⋅m^2/C.

The Electric Field in the Region of Space Shown Is