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 (2, 1, 0), (0, 1, -1), and (1, 0, 2). The surface intersects with an electric field given by 4.5 * 10⁵ $( 3 \boldsymbol { i } + 12 \boldsymbol { j } + 3 \boldsymbol { k } )$ N/C. 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 (2, 1, 0), (0, 1, -1), and (1, 0, 2), which is given by the vector (2, -1, 1). The dot product of this vector with the electric field is (2, -1, 1) · (3, 12, 3) = 6 - 12 + 3 = -3. The electric flux is then given by the product of the dot product and the area of the surface, which is (-3) * 0.20 m^2 = -0.60 N.m^2/C. However, the electric flux is defined as the amount of electric field passing through the surface, so the negative sign indicates that the electric field is passing through the surface in the opposite direction of the normal. Therefore, the electric flux through the surface is 0.60 N.m^2/C.

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

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