A cube has sides of length L = 0.170 m. It is placed with one corner at the origin as shown in the figure below. The electric field is not uniform but is given by = (-6.00 N/C · m)x + (2.00 N/C · m)z.


(a) Find the electric flux through each of the six cube faces S1, S2, S3, S4, S5, and S6.

Flux=E dot area.

It is not clear to me where E direction vectors are in your equation.

To find the electric flux through each of the six cube faces, you can use Gauss's Law, which states that the electric flux through a closed surface is directly proportional to the charge enclosed by that surface.

Here's how you can calculate the electric flux through each face:

1. Face S1: This face is perpendicular to the x-axis and its normal vector points in the positive x-direction. To calculate the electric flux through S1, we can use the formula: Φ1 = E1 ⋅ A1, where E1 is the electric field vector and A1 is the area of face S1. The area of a face of a cube is given by A = L², so the area of S1 is A1 = L². Plugging in the values, Φ1 = E1 ⋅ A1 = (-6.00 N/C · m) ⋅ (0.170 m)².

2. Face S2: This face is perpendicular to the y-axis and its normal vector points in the positive y-direction. To calculate the electric flux through S2, we can use the same formula: Φ2 = E2 ⋅ A2, where E2 is the electric field vector and A2 is the area of face S2. Since the electric field vector E2 is zero in this case (there is no y-component of the electric field), the electric flux through S2 is also zero: Φ2 = E2 ⋅ A2 = 0.

3. Face S3: This face is perpendicular to the z-axis and its normal vector points in the positive z-direction. To calculate the electric flux through S3, we can use the formula: Φ3 = E3 ⋅ A3, where E3 is the electric field vector and A3 is the area of face S3. The area of S3 is A3 = L². Plugging in the values, Φ3 = E3 ⋅ A3 = (2.00 N/C · m) ⋅ (0.170 m)².

4. Faces S4, S5, and S6: These faces are opposite to faces S1, S2, and S3 respectively. Since the electric field is not directed toward these faces, the electric flux through these faces is zero: Φ4 = Φ5 = Φ6 = 0.

Therefore, the electric flux through each face of the cube is:

Φ1 = (-6.00 N/C · m) ⋅ (0.170 m)²
Φ2 = 0
Φ3 = (2.00 N/C · m) ⋅ (0.170 m)²
Φ4 = Φ5 = Φ6 = 0

You can now calculate the values of Φ1 and Φ3 by substituting the given values into the equations.