Final answer:
The electric flux through one face of the cube can be determined using Gauss's Law. It is given by the product of the electric field and the area of the face. In this case, the total electric flux through one face of the cube is 7.272 x 10^11 Nm^2/C.
Step-by-step explanation:
The electric flux through one face of the cube can be determined by applying Gauss's Law. The electric field due to the charged particle at the center of the cube will have equal magnitude and point radially outward or inward from each face of the cube. Therefore, the electric flux through one face of the cube is given by the product of the electric field and the area of the face, and can be calculated using the formula φ = EA, where φ is the electric flux, E is the electric field, and A is the area of the face.
In this case, there are six faces of the cube, and the electric field due to the central particle is given by the formula E = kQ/r^2, where k is the electric constant, Q is the charge, and r is the distance from the center of the cube to the face. Assuming that the charges are evenly distributed on each face, the area of each face is L^2, where L is the length of the edge of the cube. Therefore, the total electric flux through one face of the cube is 6EA = 6(kQ/r^2)(L^2).
Substituting the given values, Q = 5.00 C, L = 0.140 m, k = 8.99 x 10^9 Nm^2/C^2, and r = L/2 = 0.070 m:
φ = 6(8.99 x 10^9 Nm^2/C^2)(5.00 C)/(0.070 m)^2 = 7.272 x 10^11 Nm^2/C.