A flat surface having an area of 2.90 m2 is placed in various orientations in a

Question

A flat surface having an area of 2.90 m2 is placed in various orientations in a uniform electric field of magnitude E = 5.30 x 105 N/C. (a) Calculate the electric flux through this area when the electric field is perpendicular to the surface. Apply the definition of electric flux, taking account of the angle between the field and the surface. Note that the normal to the surface is perpendicular to the plane of the surface. Consider the cosine of the angle between the electric field direction and the normal to the surface. (b) Calculate the electric flux through this area when the electric field is parallel to the surface. DE N-m2/ C (c) Calculate the electric flux through this area when the electric field makes an angle of 64.0 with the plane of the surface. You appear to be interchanging the angle the field makes with the surface and the angle between the fleld and the normal to the surface.

Explanation / Answer

Here,

A = 2.90 m^2

E = 5.3 *10^5 N/C

a) electric flux = E * A

electric flux = 5.3 *10^5 * 2.90

electric flux = 1.54 *10^6 N.m^2/C

the electric flux is 1.54 *10^6 N.m^2/C

b) fir field parallel to surface

angle between normal and field , theta = 90 degree

electric flux = E * A * cos(theta)

as cos(90 degree) = 0

electric flux = 0

c) for angle = 64 degree with surface

angle = 90 - 64 = 26 degree between electric field and normal to surface

electric flux = 5.3 *10^5 * 2.90 * cos(26 degree)

electric flux = 1.38 *10^6 N.m^2/C

the electric flux is 1.38 *10^6 N.m^2/C

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