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bija089 [108]
4 years ago
10

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 b

etween the electric field vector and the vector normal to the rectangular plane is 65.0°. What is the electric flux through the rectangle? A) 1.56 × 106 N⋅m2/C B) 6.60 × 105 N⋅m2/C C) 1.42 × 105 N⋅m2/C D) 5.49 × 104 N⋅m2/C E) 4.23 × 104 N⋅m2/C
Physics
1 answer:
jeyben [28]4 years ago
8 0

Answer:

6.60\cdot 10^5 Nm^2/C

Explanation:

The electric flux through the rectangle is given by

\Phi = E A cos \theta

where

E is the electric field strength

A is the area of the rectange

\theta is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle

In this problem we have

E = 125 000 N/C

The area of the rectangle is

A=2.50 m \cdot 5.00 m=12.5 m^2

and the angle is

\theta=65.0^{\circ}

so, the electric flux is

\Phi = (125,000 N/C)(12.5 m^2)(cos 65^{\circ})=6.60\cdot 10^5 Nm^2/C

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