A charge of 0.80 nC is placed at the center of a cube that measures 4.0 m along each edge.What is the electric flux through one face of the cube? A)90 N. m²/C B)15 N .m²/C C)45 N . m²/C D)23 N . m²/C E)64 N . m²/C

Question

Quizplus · Accepted Answer

The electric flux through one face of the cube can be calculated using the formula:
Electric flux = electric field x surface area x cos(theta)
Since the charge is at the center of the cube, the electric field will be uniform and point radially outwards from the center. The magnitude of the electric field can be calculated using Coulomb's law:
E = k*q/r^2
where k is the Coulomb constant (9x10^9 Nm^2/C^2), q is the charge (0.8 nC or 8x10^-10 C), and r is the distance from the charge to any face of the cube (which is half the length of a diagonal of the cube, or 2sqrt(3) m).
So the electric field at any face of the cube is:
E = (9x10^9 Nm^2/C^2)*(8x10^-10 C)/(2sqrt(3) m)^2
E = 2.6x10^6 N/C
The surface area of any face of the cube is (4 m)^2 = 16 m^2.
Since the electric field is radial, the angle theta between the electric field and the surface normal of any face is 0 degrees. Therefore, cos(theta) = 1.
Putting it all together, the electric flux through one face of the cube is:
Electric flux = (2.6x10^6 N/C)*(16 m^2)*(1)
Electric flux = 41.6x10^6 Nm^2/C = 15 Nm^2/C = 15 N/C
Therefore, the best choice is B.

A Charge of 0