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A square of 3.2mm on each side os immersed in a uniform e_field of magnitude E=18000N/C which makes an angel of 35° with the normal line, what's the electric flux through the surface?

A. 0.0 N·m²/C
B. 0.6 N·m²/C
C. 0.9 N·m²/C
D. 1.2 N·m²/C

User Yqritc
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1 Answer

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Final answer:

The electric flux through a square of 3.2 mm on each side in a uniform electric field of magnitude E=18000 N/C at an angle of 35° with the normal line is 0.9 N·m²/C.

Step-by-step explanation:

To find the electric flux through the surface of a square immersed in a uniform electric field, you can use the equation Φ = E * A * cos(θ), where Φ is the electric flux, E is the magnitude of the electric field, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface.

The square has a side length of 3.2 mm (which is 3.2 x 10-3 meters), so the area A = (3.2 x 10-3 m)2. The electric field E is given as 18000 N/C. The angle θ is 35° from the normal. Therefore, the electric flux Φ = 18000 N/C * (3.2 x 10-3 m)2 * cos(35°).

Calculating the cosine of 35° and then calculating Φ, we find:
Φ = 18000 * (3.2 x 10-3)2 * cos(35°).
Φ = 18000 * (1.024 x 10-5) * 0.8192.
Φ = 0.9 N·m2/C.

Therefore, the correct answer is C. 0.9 N·m2/C.

User Eray Balkanli
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