Answer:
6.07 Nm²/C
Explanation:
L = length of the flat sheet = 0.400 m
w = width of the flat sheet = 0.600 m
Area of the rectangular sheet is given as
A = L w
A = (0.400) (0.600)
A = 0.24 m²
E = magnitude of the electric field in the region = 73.9 N/C
θ = Angle of the electric field relative to normal to the plane of sheet = 90 - 20 = 70 deg
Magnitude of magnetic flux through the sheet can be given as
Φ = E A Cosθ
Inserting the values
Φ = (73.9) (0.24) Cos70
Φ = 6.07 Nm²/C
This question involves the concepts of the electric flux and the electric field.
The magnitude of electric flux is found to be "6.1 N.m²/C".
The electric flux can be given by the following formula:
[tex]\phi = EACos\theta[/tex]
where,
[tex]\phi[/tex] = electric flux = ?
E = Electric field = 73.9 N/C
A = Area of the plate = (0.4 m)(0.6 m) = 0.24 m²
θ = angle between normal to area and electric field lines = 90° - 20° = 70°
Therefore,
[tex]\phi=(73.9\ N/C)(0.24\ m^2)Cos(70^o)\\\phi = 6.1\ N.m^2/C[/tex]
Learn more about electric flux here:
https://brainly.com/question/14850656?referrer=searchResults
The attached picture shows the electric flux.
