what is the solution to these questions? An electric field of intensity 4.1 kN/C
ID: 1675505 • Letter: W
Question
what is the solution to these questions? An electric field of intensity 4.1 kN/C is applied along the x-axis. Calculatethe electric flux through a rectangular plane 0.350 m wide and0.700 m long if the following conditions are true. (a) The plane is parallel to the yz-plane.Nm2/C
(b) The plane is parallel to the xy-plane.
Nm2/C
(c) The plane contains the y-axis, and its normal makes an angle of50.0° with the x-axis.
Nm2/C
what is the solution to these questions? An electric field of intensity 4.1 kN/C is applied along the x-axis. Calculatethe electric flux through a rectangular plane 0.350 m wide and0.700 m long if the following conditions are true. (a) The plane is parallel to the yz-plane.
Nm2/C
(b) The plane is parallel to the xy-plane.
Nm2/C
(c) The plane contains the y-axis, and its normal makes an angle of50.0° with the x-axis.
Nm2/C
(a) The plane is parallel to the yz-plane.
Nm2/C
(b) The plane is parallel to the xy-plane.
Nm2/C
(c) The plane contains the y-axis, and its normal makes an angle of50.0° with the x-axis.
Nm2/C
Explanation / Answer
= E. A Given E = 4.1kN/C along X axis = 4100i^ N/C (i^is unit vector along X axis) Area A = 0.350* 0.700 = 0.245 m2 A = A* n^ here A isarea ( in vecor form), n^ is normal unit vector i.e. unit vectornormal to the surface. a. for A in YZplane, n^ = i^ => A = 0.245i^ yz = 4100i^ . 0.245 i^ = 1004.5 kN-m2/C (as i^ .i^ = 1) => A = 0.245k^ yz = 4100i^ . 0.245 k^ = 0 (as i^ . k^ = 0) c. = E. A = E* A * cos given = 50.00 50.0 = 4100* 0.245 * cos 50.00 = 0.646 kN-m2/ C = 646 N-m2/ C = 0.646 kN-m2/ C = 646 N-m2/ CRelated Questions
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