If a charge of +Q is placed at one corner of a cube, what is the total electric

If a charge of +Q is placed at one corner of a cube, what is the total electric flux through the entire surface of the cube?

If a charge of +Q is placed at one corner of a cube, what is the total electric flux through the entire surface of the cube?

Step 1: Understand that electric flux is related to the charge enclosed by a surface.
Step 2: Recall Gauss's Law, which states that the total electric flux (Φ) through a closed surface is equal to the charge (Q) enclosed divided by the permittivity of free space (ε₀).
Step 3: In this case, we have a charge of +Q placed at one corner of a cube.
Step 4: Visualize the cube: since the charge is at a corner, only a fraction of the charge is actually inside the cube.
Step 5: Determine how much of the charge is inside the cube: since the charge is at a corner, only 1/8 of the charge would be inside if the cube were divided into 8 smaller cubes.
Step 6: However, since we are considering the entire cube and the charge is at the corner, we need to consider that the cube shares this corner with 6 other cubes.
Step 7: Therefore, the fraction of the charge that is effectively enclosed by the cube is 1/6.
Step 8: Now, apply Gauss's Law: the total electric flux through the cube is Φ = (Q/6) / ε₀.
Step 9: Simplify the expression: Φ = Q / (6ε₀).

Gauss's Law – The total electric flux through a closed surface is proportional to the charge enclosed within that surface.
Symmetry in Charge Distribution – Understanding how the placement of a charge affects the distribution of electric field lines and flux through surfaces.