ok the question is....A point charge of 8.00 nC is at the center of a cube with sides that are 0.200 m. What is the electric flux through (a) the surface of the cube, (b) one of the six faces of the cube?
ok so would I just use Gauss's law 4* pi*k*q? even though that is spherical? I am confused on how to determine what equation to use for the whole surface and for just one side
(2) Yes, use Gauss' law. It is good for any closed surface shape surrounding a charge, and does not depend upon the location of the charge!
In your case, since the charge is in the center, just divide by six to get the flux through one side, because of the symmetry of the cube and the charge's location.
ok so the length of the sides doesn't matter in this case? I am confused! haha
Yes, the cube length does not matter for the FLUX. The E-field will decrease for larger cubes, but the area increase will cancel it out.
To determine the electric flux through the surface of the cube and through one of its faces, you need to use Gauss's law. However, you cannot directly apply the equation for a point charge in a spherical Gaussian surface to a cube.
Here's how you can determine the electric flux through the surface of the cube and through one face using Gauss's law:
(a) Electric flux through the surface of the cube:
To find the electric flux through the entire surface of the cube, you can imagine a Gaussian surface enclosing the cube. In this case, a cube is a symmetrical object, so you can choose a cube as your Gaussian surface.
Gauss's law states that the electric flux through a closed surface is directly proportional to the charge enclosed by the surface. Mathematically, it can be expressed as:
Φ = Q_enclosed / ε₀
where Φ is the electric flux, Q_enclosed is the charge enclosed by the Gaussian surface, and ε₀ is the permittivity of free space.
In this case, the charge enclosed by the Gaussian surface is 8.00 nC (8.00 × 10^-9 C). Thus, the electric flux through the surface of the cube is:
Φ = Q_enclosed / ε₀ = (8.00 × 10^-9 C) / (8.854 × 10^-12 C^2 / N m^2)
Substituting the values and calculating the result will give you the electric flux through the surface of the cube.
(b) Electric flux through one face of the cube:
To find the electric flux through a single face of the cube, you need to consider a smaller Gaussian surface corresponding to that specific face. Since the cube has six faces, each face will have the same magnitude of electric flux.
Again, use Gauss's law and consider a Gaussian surface in the shape of a square, with one face of the cube as its base. The area of the square face will act as a factor to scale down the electric flux.
The electric flux through one face of the cube can be calculated by dividing the electric flux through the entire surface of the cube by the number of faces. Therefore:
Flux through one face = Flux through the surface of the cube / Number of faces
Substitute the value calculated in part (a) for the electric flux through the surface of the cube and divide it by 6 to obtain the electric flux through one face of the cube.
Remember, when using Gauss's law, it is essential to choose an appropriate Gaussian surface that exhibits the same symmetry as the given charge distribution.