A large charge distribution Q exists at the center of a cube. It is surrounded by six smaller charges, q , each facing a direction perpendicular to a face of the cube.What is the electric flux through one of the sides of the cube?
To find the electric flux through one side of the cube, we can use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε₀).
In this case, the cube has six sides, so we need to find the electric flux through one of those sides. We'll assume that one side of the cube is a Gaussian surface.
To apply Gauss's Law, we need to determine the total charge enclosed by the Gaussian surface. In this scenario, the large charge distribution Q at the center of the cube is enclosed by the surface, while the smaller charges q are outside the surface. Since the flux through a closed surface due to external charges is zero, we can ignore the smaller charges q.
Therefore, the total charge enclosed by the Gaussian surface is Q.
Now we can calculate the electric flux (Φ) through one side of the cube using Gauss's Law:
Φ = Q/ε₀
where ε₀ is the permittivity of free space.
It's important to note that the direction of the electric field and flux is perpendicular to the surface of the cube due to the symmetry of the problem.
So, to find the electric flux through one of the sides of the cube, divide the total charge enclosed (Q) by the permittivity of free space (ε₀).