A flat 1.0 m2 surface is vertical at
x = 2.0 m and parallel to the yz-plane. What is the flux through the surface if it is located in a uniform electric field given by E = 25.0i + 42.0j + 62.0k N/C?
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To calculate the flux through a surface in an electric field, you need to use Gauss's law, which states that the electric flux through a closed surface is directly proportional to the electric charge enclosed by that surface.
In this case, we are given a flat surface with an area of 1.0 m^2 that is vertical and parallel to the yz-plane. Let's call this surface S.
First, we need to calculate the angle between the surface normal and the electric field. Since the surface is vertical, the surface normal is perpendicular to the x-direction. Therefore, the angle between the surface and electric field is 90 degrees.
Now, we can calculate the electric flux using the formula: Flux = E * A * cos(theta), where E is the electric field vector, A is the area vector, and theta is the angle between them.
The electric field vector is given as E = 25.0i + 42.0j + 62.0k N/C.
The area vector can be calculated using the cross product of two vectors parallel to the surface. Since the surface is parallel to the yz-plane, the vectors parallel to the surface are i.
So, the area vector A = i.
The angle between the electric field and the area vector is 90 degrees in this case.
Plugging these values into the electric flux formula, we get:
Flux = (25.0i + 42.0j + 62.0k) * i * cos(90°)
Since cos(90°) = 0, the y and z components of the electric field are not contributing to the flux in this case.
Flux = (25.0i) * (1.0i) * 0 = 0
Therefore, the flux through the surface S is 0 N*m^2/C.