Electric flux through one face of the cube.

[musicman05]

A point charge Q = 5.00 µC is located at the center of a cube of side L= 0.120 m. In addition, six other identical point charges having q = -0.50 µC are positioned symmetrically around Q, as shown in Figure P24.19. Determine the electric flux through one face of the cube.

 

[Snazzy]

Okay, so what have you done so far?

 

[pam]

Use Gauss's law and the symmetry of a cube.

 

[human_eraser]

same sort of question, is L not needed for the calculation?
can someone check my work,
Q=5.17 uC
q=-1.10 uC
L= 0.124 m
Qnet=Q+(1/6)q
flux=Qnet/epsilon
per side=flux/6
26900 N*m^2/C?

 

[paranormal]

The electric flux through one face of the cube can be calculated using the formula Φ = ∫E⋅dA, where Φ is the electric flux, E is the electric field, and dA is the differential area. In this case, the electric field can be found by superimposing the electric fields of all seven point charges. Using the principle of superposition, we can see that the electric field at any point on the face of the cube will be the sum of the electric fields due to each individual point charge.

To find the electric field at a point on the face of the cube, we can use Coulomb's law, which states that the electric field at a point due to a point charge is given by E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the point charge, and r is the distance between the point charge and the point where the electric field is being measured.

In this case, the electric field at a point on the face of the cube due to the central point charge Q will be E = kQ/r^2, where r is the distance from the center of the cube to the face, which is L/2. Thus, the electric field due to Q at a point on the face of the cube will be E = kQ/(L/2)^2 = 4kQ/L^2.

Similarly, the electric field due to each of the six -0.50 µC point charges at a point on the face of the cube can be found using the same formula, with r being the distance from the point charge to the face, which is also L/2. Thus, the electric field due to each of these six point charges will be E = -4kq/L^2.

Since the electric field is a vector quantity, we need to take into account the direction of the electric fields due to each point charge. In this case, the electric fields due to the six -0.50 µC point charges will point away from the face of the cube, while the electric field due to the central point charge Q will point towards the face of the cube.

Thus, the total electric field at a point on the face of the cube will be the sum of the electric fields due to all seven point charges, which can be written as E = 4k(Q - 6q)/L^2.

Substituting the