An constant electric field, $$\overrightarrow { \mathbf { E } } = ( 3 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } } )$$ N/C,goes through a surface with area $$\overrightarrow { \mathbf { A } } = ( 8 \hat { \mathbf { i } } - 6 \hat { \mathbf { k } } )$$ m².(This surface can also be expressed as an area of 10 m² with the direction of the unit vector ( $0.8 \tilde { \mathbf { i } } - 0.6 \tilde { \mathbf { k } }$ ).What is the magnitude of the electric flux through this area?

Question

Quizplus · Accepted Answer

The electric flux through a surface is given by the dot product of the electric field and the surface area vector. In this case, we have:$$\overrightarrow { \mathbf { \Phi } } = \overrightarrow { \mathbf { E } } \cdot \overrightarrow { \mathbf { A } } = ( 3 	ilde { \mathbf { i } } + 4 \hat { \mathbf { j } } ) \cdot ( 8 \hat { \mathbf { i } } - 6 \hat { \mathbf { k } } ) = 24 	ext{ N m}^2/	ext{C}$$Therefore, the magnitude of the electric flux through the surface is 24 N m2/C.

An Constant Electric Field E→=(3i~+4j^)\overrightarrow { \mathbf { E } } = ( 3 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } } )E=(3i~+4j^​)

An Constant Electric Field $\overrightarrow { \mathbf { E } } = ( 3 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } } )$