The surface shown is a square and has a side length of 14.0 cm. The electric field acting on this area has a magnitude of 180. N/C at an angle of 27.5°. Calculate the electric flux through the shown surface.
angle measured how? to surface, or the Normal?
Assuming to the normal
flux=E*Area*cosTheta
its to the surface
then use sin Theta
To calculate the electric flux through a surface, you can use the formula:
Flux = Electric Field * Surface Area * Cos(θ)
where:
- Flux is the electric flux
- Electric Field is the magnitude of the electric field
- Surface Area is the area of the surface
- Cos(θ) is the cosine of the angle formed between the electric field vector and the surface normal vector.
In this case, you are given:
- Electric Field = 180 N/C (magnitude of the electric field)
- Surface Area = side length * side length = 14.0 cm * 14.0 cm (since the surface is a square)
- θ = 27.5° (angle between the electric field vector and the surface normal vector)
Let's plug these values into the formula:
Flux = 180 N/C * (14.0 cm * 14.0 cm) * Cos(27.5°)
Before calculating, let's convert the side length from cm to meters, since the SI unit for electric field is N/C (newtons per coulomb) and the SI unit for area is m^2 (square meters).
1 cm = 0.01 m
Flux = 180 N/C * (0.14 m * 0.14 m) * Cos(27.5°)
Now we can calculate the flux:
Flux = 180 N/C * 0.0196 m^2 * Cos(27.5°)
Flux ≈ 46.73 N * m^2 / C
Therefore, the electric flux through the shown surface is approximately 46.73 N * m^2 / C.