A flat rectangular surface (1 = 4.6 mm and b =2.0 mm) lies in the horizontal pla

Question

A flat rectangular surface (1 = 4.6 mm and b =2.0 mm) lies in the horizontal plane in a region of a uniform electric field of 1400 N/C. The electric field makes an angle alpha = 42.0 degree with the horizontal. For simplicity, only one field line is shown in the diagram below. What is the electric flux through the surface if the normal to the surface is directed vertically up? N m^2/c What is the electric flux through the surface if the normal to the surface is directed vertically down? N. m^2/C

Explanation / Answer

(a)A = area = a x b = (4.6 x 10-3) (2 x 10-3) = 9.2 x 10-6 m2

angle with normal= 90-42=48

flux= EAcos48

Flux=1400(9.2x10-6)cos48

Flux=8.62x10-3 Nm2/C

(b) angle with normal= 90+42=132o

flux= EAcos132

Flux=1400(9.2x10-6)cos132

Flux=-8.62x10-3 Nm2/C

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