The electric field E_1 at one face of a parallelepiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field E_2 is also uniform over the entire face and is directed into that face (the figure ). The two faces in question are inclined at 30.0 degrees from the horizontal, while E_1 and E_2 are both horizontal; E_1 has a magnitude of 2.90×10^4 N/C, and E_2 has a magnitude of 8.60×10^4 N/C.

Assuming that no other electric field lines cross the surfaces of the parallelepiped, determine the net charge contained within.

Use Gauss' Law.

The sum of the fluxes (E-fields NORMAL TO the surface multiplied by the areas) for all six surfaces is proportional to the charge inside, Q.

(Total flux = Q/epsilono)

-6.93*10^-5 N/C

To solve this problem, we can use Gauss's Law, which states that the net electric flux passing through a closed surface is equal to the total charge enclosed by that surface divided by the electric constant.

1. First, let's calculate the electric flux passing through the two faces of the parallelepiped. The electric flux passing through each individual face is given by:

Flux = E * A * cos(theta)

Where E is the magnitude of the electric field, A is the area of the face, and theta is the angle between the electric field and the normal to the face.

2. The angle between the electric field and the normal to the face is 30 degrees for both faces.

3. The area of each face can be calculated as A = length * width, where length and width are the dimensions of the face (assuming the parallelepiped is rectangular).

4. The total flux passing through the two faces can be calculated by adding the flux through each face.

Total Flux = Flux1 + Flux2

5. The net charge contained within the parallelepiped is then given by:

Charge = (Total Flux) * (electric constant)

6. Let's plug in the given values:

Flux1 = (2.90x10^4 N/C) * (length * width) * cos(30 degrees)
Flux2 = (8.60x10^4 N/C) * (length * width) * cos(30 degrees)

Total Flux = Flux1 + Flux2

Charge = (Total Flux) * (electric constant)

7. Calculating the areas and adding the flux:

A = (length * width) = (length^2) * sin(60 degrees)
A1 + A2 = 2 * (length^2) * sin(60 degrees)

Flux = ((2.90x10^4 N/C) + (8.60x10^4 N/C)) * 2 * (length^2) * sin(60 degrees)

8. Multiplying by the electric constant:

Charge = Flux * (electric constant)

9. Substitute the values of electric constant:

Charge = Flux * 8.99x10^9 N m^2/C^2

10. Finally, calculate the net charge contained within the parallelepiped.

To determine the net charge contained within the parallelepiped, we can use Gauss's Law. Gauss's Law relates the electric field on a closed surface to the charge enclosed by that surface.

1. Determine the area of the inclined faces:
The inclined faces are inclined at 30 degrees from the horizontal. Assuming the parallelepiped is a rectangle, we can calculate the area of one inclined face using the formula for the area of a rectangle:
Area = base * height
In this case, the base is the length of one side of the parallelepiped, and the height is the vertical height of the parallelepiped.

2. Calculate the flux through each inclined face:
The flux through a surface is given by the product of the electric field and the area of the surface. Since the electric field is uniform and directed perpendicular to the inclined faces, the flux through each face is simply the product of the magnitude of the electric field and the area of the face.

Flux = Electric Field * Area

3. Determine the net charge enclosed by each inclined face:
The net charge enclosed by each inclined face is given by the flux through that face divided by the electric field magnitude on that face. This is based on Gauss's Law, which states that the net charge enclosed is equal to the total flux divided by the electric field.

Net Charge Enclosed = Flux / Electric Field

4. Find the net charge contained within the parallelepiped:
Since the inclined faces have opposite electric field directions, the net charge enclosed by one face will be positive, while the net charge enclosed by the other face will be negative. To find the net charge contained within the parallelepiped, we simply subtract the absolute values of the net charges on each inclined face.

Net Charge Contained = |Net Charge Enclosed by Face 1| - |Net Charge Enclosed by Face 2|