Question

a point charge of magnitude q is at the center of a cube with sides of length l. part a what is the electric flux φ through each of the six faces of the cube? use ε₀ for the permittivity of free space (not the emf symbol e₀). view available hint(s) φ = input box submit part b what would be the flux φ₁ through a face of the cube if its sides were of length l₁? use ε₀ for the permittivity of free space. view available hint(s)

Explanation:

Part A

Step1: Apply Gauss's Law

Gauss's Law states that the total electric flux through a closed surface is $\frac{q}{\epsilon_0}$, where $q$ is the enclosed charge and $\epsilon_0$ is the permittivity of free space. For a cube with a point charge at its center, the cube is a closed surface enclosing the charge $q$.

Step2: Distribute Flux Among Faces

Since the cube is symmetric and the charge is at the center, the electric flux is equally distributed among the six faces. So, the flux through each face is the total flux divided by 6. The total flux is $\frac{q}{\epsilon_0}$, so dividing by 6 gives $\frac{q}{6\epsilon_0}$.

Step1: Recall Gauss's Law and Symmetry

Gauss's Law depends on the enclosed charge, not the size of the closed surface. The point charge $q$ is still enclosed by the cube (even with side length $L_1$), so the total electric flux through the cube remains $\frac{q}{\epsilon_0}$.

Step2: Distribute Flux Among Faces

Due to the symmetry of the cube (the charge is at the center, so each face is equivalent), the flux through each face is still the total flux divided by 6. So, the flux through one face is $\frac{q}{6\epsilon_0}$, same as in Part A, because the enclosed charge and the symmetry (6 faces) don't change with the side length of the cube.

Answer:

$\frac{q}{6\epsilon_0}$

Part B