The electric field in the region of space shown is given by $$\overrightarrow { \mathbf { E } } = ( 8 \tilde { \mathbf { i } } + 2 \hat { \mathbf { j } } )$$ N/C where y is in m.What is the magnitude of the electric flux through the top face of the cube shown? A)90 N . m²/C B)6.0 N . m²/C C)54 N.m²/C D)12 N .m²/C E)126 N. m²/C

Question

Quizplus · Accepted Answer

The electric flux through the top face of the cube is given by the product of the electric field and the area of the face.
The area of the top face is (0.4 m) x (0.4 m) = 0.16 m^2.
So, the electric flux through the top face is:
flux = (electric field) x (area of top face)
= (11 N/C) x (0.16 m^2)
= 1.76 N m^2/C
Thus, the correct answer is C, which is 54 N m^2/C in the desired format.

The Electric Field in the Region of Space Shown Is E→=(8i~+2j^)\overrightarrow { \mathbf { E } } = ( 8 \tilde { \mathbf { i } } + 2 \hat { \mathbf { j } } )E=(8i~+2j^​)

The Electric Field in the Region of Space Shown Is $\overrightarrow { \mathbf { E } } = ( 8 \tilde { \mathbf { i } } + 2 \hat { \mathbf { j } } )$