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A flat surface of area 3.20 m2 is rotated in a uniform electric field of magnitude e = 6.20 x 105 n/c. determine the electric flux through this area (a) when the electric field is perpendicular to the surface and (b) when the electric field is parallel to the surface.

Respuesta :

The electric flux on a surface of area A in a uniform electric field E is given by 
φ = EA cos θ
where θ = the angle between the directions of E and a normal vector to the surface of the area.
See the diagram shown below.

When the electric field is perpendicular to the surface, then θ = 0°, and
φ = EA cos(0°)  = (6.20 x 10⁵ N/C)*(3.2 m²) = 1.984 x 10⁶ (N-m²)/C

When the electric field is parallel to the area, then θ = 90°, and
φ = EA cos(90°) = 0

Answer:
(a) 1.984 x 10⁶ (N-m²)/C
(b) 0
Ver imagen Аноним
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The electric flux through this area:

  • When the electric field is perpendicular to the surface= 1.98 x 10⁶ Nm²/C
  • When the electric field is parallel to the surface= 0

What is Electric flux?

This is defined as the number of electric field lines that intersect a given area.

Electric flux(φ)= EA cos θ

where θ = angle between the directions of E and A is the surface area.

When the electric field is perpendicular to the surface,  θ = 0°.

φ = EA cos(0°)  = (6.20 x 10⁵ N/C)*(3.2 m²) × 1 = 1.984 x 10⁶ Nm²/C

When the electric field is parallel to the area, then θ = 90°, and

φ = EA cos(90°) = 0 as result of cos 90° being zero.

Read more about Electric flux here https://brainly.com/question/26289097

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