.-/1.66 points SerCP11 15.P.045.MI My Notes An electric field of intensity 2.85
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.-/1.66 points SerCP11 15.P.045.MI My Notes An electric field of intensity 2.85 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long if the following conditions are true. (a) The plane is parallel to the yz-plane. N m2/C (b) The plane is parallel to the xy-plane. N m2/c (c) The plane contains the y-axis, and its normal makes an angle of 33.0° with the x-axis. N m2/c Need Help? 1.7/1.7 points 1 Previous Answers SerCP11 15.P057 My NotesExplanation / Answer
When the field is uniform over the rectangular area the electric flux is E x area (otherwise it is integral E.ds)
a) The area is normal to E
Flux = 2.85 x10^3 x (0.35 x0.7) = 698.25 N/C/m^2
b) As the area is parallel to the field no field lines through it.
Flux = 0
c)The area makes an angle of 45 deg with the x axis. The component of E through the area is E cos 33
Flux = E cos 33 x area
........= 585.60 N/C/m^2
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