FIGURE 19-15
-A uniform electric field with a magnitude of 6 × 10⁶ N/C is applied to a cube of edge length 0.1 m as shown in Figure 19-15. If the direction of the E-field is along the +x-axis, what is the electric flux passing through the shaded face of the cube?

Question

FIGURE 19-15

-A uniform electric field with a magnitude of 6 × 10⁶ N/C is applied to a cube of edge length 0.1 m as shown in Figure 19-15. If the direction of the E-field is along the +x-axis, what is the electric flux passing through the shaded face of the cube?

Quizplus · Accepted Answer

The electric flux through a surface is given by the formula Φ = E * A * cos(θ), where E is the electric field magnitude, A is the area of the surface, and θ is the angle between the field and the normal to the surface. For the shaded face of the cube, A = (0.1 m)^2 = 0.01 m², and θ = 0° since the field is along the +x-axis and the normal to the shaded face is also along the +x-axis. Thus, Φ = 6 × 10^6 N/C * 0.01 m² * cos(0°) = 6 × 10^4 Nm²/C.

FIGURE 19-15 -A Uniform Electric Field with a Magnitude of 6 ×