A flat surface having an area of 3.2 m2 is rotated in a uniform electric field o

Question

A flat surface having an area of 3.2 m2 is rotated in a uniform electric field of magnitude E = 5.8 105 N/C. Determine the electric flux through this area in the following situations: (a) when the electric field is perpendicular to the surface N

Explanation / Answer

electric flux through this area = phi (e) = dot product of electric field vector and normal area vector phi (e) = E (dot) dS phi (e) = |E|* |ds| cos ? where (?) is the angle, E vector makes with normal area vector 1) when electric field is perpendicular to the surface ? =0 {if roof is surface, then an arrow from floor through roof is the area vector >> when E is also from floor through roof, E is perpendicular to crossetion of area >> or parallel to area vector phi (e) = 5.8*10^5* 3.2 cos o = 18.56*10^5 (N-m^2/C) 1) when electric field is parallel to the surface ? = 90 deg E parallel to crosection of surface E and area vector become perpendicular no flux passes through phi (e) = 5.8*10^5* 3.2 cos 90 = 0

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