Question

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? 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

Answer 1

`6.60\cdot 10^5 Nm^2/C`

Step-by-step 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`