A surface that has an area of 0.20 m² lies on a plane whose normal is perpendicular to the plane defined by the points (3, 0, 0), (0, 1, 0), and (0, 0, 3). The surface intersects with an electric field given by 8.5 $\times$ 10⁵ $( 4 \boldsymbol { i } + 12 \boldsymbol { j } - 3 \boldsymbol { k } )$ N/C. The magnitude of the electric flux through the surface is

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Quizplus · Accepted Answer

The electric flux through a surface is given by the dot product of the electric field and the normal to the surface. In this case, the normal to the surface is perpendicular to the plane defined by the points (3, 0, 0), (0, 1, 0), and (0, 0, 3), which is $\boldsymbol { n } = \frac { 1 } { \sqrt { 14 } } ( 3 \boldsymbol { i } + \boldsymbol { j } + 3 \boldsymbol { k } )$. The electric field is given by $\boldsymbol { E } = 8.5 \times 10^5 ( 4 \boldsymbol { i } + 12 \boldsymbol { j } - 3 \boldsymbol { k } )$ N/C. Therefore, the electric flux through the surface is:$$\Phi_e = \boldsymbol { E } \cdot \boldsymbol { n } A = 8.5 \times 10^5 \left( \frac { 1 } { \sqrt { 14 } } ( 3 \boldsymbol { i } + \boldsymbol { j } + 3 \boldsymbol { k } ) \right) \cdot ( 4 \boldsymbol { i } + 12 \boldsymbol { j } - 3 \boldsymbol { k } ) (0.20 \text{ m}^2)$$$$= 8.5 \times 10^5 \left( \frac { 1 } { \sqrt { 14 } } ( 12 - 3 + 9 ) \right) (0.20 \text{ m}^2)$$$$= 1.9 \times 10^6 \text{ N.m}^2/\text{C}$$

A Surface That Has an Area of 0 ×\times× 105 (4i+12j−3k)( 4 \boldsymbol { i } + 12 \boldsymbol { j } - 3 \boldsymbol { k } )(4i+12j−3k)

A Surface That Has an Area of 0 $\times$ 10⁵ $( 4 \boldsymbol { i } + 12 \boldsymbol { j } - 3 \boldsymbol { k } )$