Therefore, the net electric flux through the piece-wise smooth surfaces that comprise the cube is approximately 4.91 x 10^9 Nm^2/C. Net electric flux = (Electric flux through one face) x (Number of faces) Therefore, the net electric flux through the piece-wise smooth surfaces that comprise the cube is: It should be noted that electric flux is defined as the number of electric field lines which are passing through a given area in a unit time. Since the cube is symmetrical, the electric flux through each face will be the same as calculated in part (a), which is approximately 8.18 x 10^8 Nm^2/C. Electric flux formula is obtained by multiplying the electric field and the component of the area perpendicular to the field.

(b) To determine the net electric flux through the piece-wise smooth surfaces that comprise the cube, we need to consider all six faces of the cube. Therefore, the electric flux through any of the faces of the cube is approximately 8.18 x 10^8 Nm^2/C. If the E-field is not perpendicular to the surface area, then the flux will be less than EA because less electric field lines will penetrate A. Since the charge is at the center of the cube, the distance from the charge to any point on the surface of the cube is half the length of the side of the cube, which is 30 cm = 0.3 m.Įlectric field = (1 / (4πε0)) x (2.0 nC / (0.3 m)^2)Įlectric field = (1 / (4π(8.85 x 10^-12 C^2/Nm^2))) x (2.0 x 10^-9 C / (0.3 m)^2)Įlectric flux = (2.27 x 10^9 N/C) x (0.36 m^2)

The electric field at any point on the surface of the cube is given by Coulomb's Law:Įlectric field = (1 / (4πε0)) x (Q / r^2) The electric flux through this face is given by the equation:Įlectric flux = (Electric field) x (Area) The area of one face of the cube is (60 cm)^2 = 3600 cm^2 = 0.36 m^2. Since the cube is symmetrical, we can choose any face to calculate the electric flux. The permittivity of free space is ε0 = 8.85 x 10^-12 C^2/Nm^2. In this case, the charge enclosed is the point charge Q = 2.0 nC. L L We need to solve for the electric flux through each face of the cube and take the sum to calculate. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε0). Calculate the electric flux through the closed surface. (a) To determine the electric flux through any of the faces of the cube, we can use Gauss's Law.

Electric flux equation cube