A uniform electric field of magnitude 2.10×10^4 makes an angle of 37 with a plane surface of area 1.60×10^−2 .What is the electric flux through the surface?
(E-field)*(Area)*sin37 = ___
newton*m^2/Coulomb
You should have provided units for your E-field, area and angle. I assumed that are N/C, square meters and degrees
To find the electric flux through the surface, you can use the formula:
Electric Flux (Φ) = Electric Field (E) * Area (A) * cosine(theta)
Given:
Electric Field (E) = 2.10×10^4
Area (A) = 1.60×10^−2
Angle (theta) = 37°
First, convert the angle from degrees to radians:
theta (in radians) = 37° * (π/180°)
Now, calculate the electric flux using the formula:
Φ = E * A * cos(theta)
Φ = (2.10×10^4) * (1.60×10^−2) * cos(37° * (π/180°))
Finally, calculate the value:
Φ ≈ 535.19
Therefore, the electric flux through the surface is approximately 535.19.