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A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle between the electric field vector and the vector normal to the rectangular plane is 65.0°. What is the electric flux through the rectangle?

Respuesta :

Answer:

6.6 x 10^5 Nm^2/C

Explanation:

E = 125000 N/C

Area, A = length x width = 2.5 x 5 = 12.5 m^2

θ = 65 degree

electric flux, φ = E A Cosθ

φ = 125000 x 12.5 x Cos 65

φ = 6.6 x 10^5 Nm^2/C

adioabiola

Here, we are required to evaluate the electric flux through the rectangle.

The electric flux through the rectangle is;

φ = 6.6 × 10^5 Nm²/C

The electric flux through a body is given as;

electric flux, φ = E A Cosθ

where;.

  • φ = electric flux
  • E = Electric field magnitude
  • A = Area
  • θ = angle between the electric field vector and the vector normal

The Area of the rectangle, A = 2.5 × 5

Therefore, A = 12.5m².

Therefore, the electric flux, φ = E A Cosθ

φ = 125000 × 12.5 × Cos65

φ = 660,341 Nm²/C.

φ = 6.6 × 10^5 Nm²/C

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