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

Question

A flat rectangular surface (a = 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 m2/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 = 4.6 mm x 2 mm = 9.2 mm^2 = 9.2 x 10^-6 m^2

E = 1400 N/C

We know that,

phi = EA cos(theta)

phi = 1400 x 9.2 x 10^-6 x cos(42) = 1.02 x 10^-2 N m^2/C

Hence, phi = 1.02 x 10^-2 N m^2/C

b)phi = EA cos(theta)

phi = 1400 x 9.2 x 10^-6 x cos(-42) = 1.02 x 10^-2 N m^2/C

Hence, phi = 1.02 x 10^-2 N m^2/C

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