Question

A point charge +q is placed precisely at the center of one of the faces of a perfectly symmetrical cube. What is the total electric flux passing through the cube?

• A. q/6\(\epsilon_0\)
• B. q/2\(\epsilon_0\)
• C. q/\(\epsilon_0\)
• D. q/3\(\epsilon_0\)

Hint:

Remember common symmetric charge placements for flux through a cube:
- Charge at center: \( q/\epsilon_0 \)
- Charge at center of a face: \( q/2\epsilon_0 \)
- Charge at midpoint of an edge: \( q/4\epsilon_0 \)
- Charge at a corner/vertex: \( q/8\epsilon_0 \)
These are derived by finding how many identical cubes are needed to symmetrically enclose the charge.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the total electric flux passing through a cube when a point charge is located exactly at the center of one of its faces. This is a problem solvable using Gauss's Law and symmetry arguments.

Step 2: Key Formula or Approach:
1. Gauss's Law: The total electric flux (\(\Phi\)) through any closed surface (Gaussian surface) is given by:
\[ \Phi = \frac{Q_{enclosed}}{\epsilon_0} \]
Where \( Q_{enclosed} \) is the net charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space.
2. Symmetry for Partial Enclosure: If a charge is not fully enclosed by a single surface, we can construct a larger, perfectly symmetrical closed Gaussian surface by adding identical imaginary surfaces until the charge is completely enclosed. The flux through the original surface will then be a fraction of the total flux through the larger symmetrical surface.

Step 3: Detailed Explanation:
1. The point charge \(+q\) is placed at the center of one face of a cube. This means the charge is on the boundary of the cube, not entirely inside it.
2. To apply Gauss's Law, we need to enclose the charge symmetrically. We can imagine placing another identical cube directly adjacent to the first cube, such that the face on which the charge rests is common to both cubes. This effectively places the charge at the center of the combined, larger cubical volume formed by the two cubes.
3. The combined structure (two cubes) now perfectly encloses the charge \(+q\).
4. According to Gauss's Law, the total electric flux through this combined Gaussian surface (made of two cubes) is:
\[ \Phi_{total} = \frac{+q}{\epsilon_0} \]
5. By symmetry, the electric flux from the charge will pass equally through the two cubes. Therefore, the flux through the single cube in question will be half of the total flux:
\[ \Phi_{cube} = \frac{1}{2} \Phi_{total} = \frac{1}{2} \frac{+q}{\epsilon_0} = \frac{q}{2\epsilon_0} \]

Step 4: Final Answer:
The total electric flux passing through the cube is q/2\(\epsilon_0\).

Answer 2

This same result can be reached through the idea of solid angle, which measures how much of the space around a point charge a given surface "covers," without needing to construct an extra imaginary cube.

1. q/6\(\epsilon_0\): This value would apply if the charge sat exactly at the centre of the cube, since by symmetry each of the six faces would then receive an equal one-sixth share of the total flux \( \frac{q}{\epsilon_0} \). Here, however, the charge sits on a face, not at the centre, so this even six-way split does not apply.
2. q/2\(\epsilon_0\): A point charge radiates its field lines equally in all directions over the full \( 4\pi \) steradians of space around it. Because the charge sits exactly on the plane of one face, the cube occupies precisely one half of all the directions the charge can radiate into, that is, a solid angle of \( 2\pi \) steradians out of the total \( 4\pi \). The flux through the cube is therefore the same fraction of the total flux: \( \Phi = \frac{2\pi}{4\pi}\times\frac{q}{\epsilon_0} = \frac{q}{2\epsilon_0} \). This matches.
3. q/\(\epsilon_0\): This would be the flux if the entire field of the charge passed through the cube, which would only be true if the cube completely enclosed the charge on all sides; since the charge lies on the boundary and half of its field lines escape to the outside region beyond that face, this option overstates the flux.
4. q/3\(\epsilon_0\): This does not correspond to any natural symmetric fraction of the charge's surroundings relative to a cube, since the charge's position on a face gives a clean half-and-half split of the surrounding space, not a one-third split.

Since exactly half of the space around the point charge lies within the cube, exactly half of its total flux of \( \frac{q}{\epsilon_0} \) passes through the cube.

Therefore, the correct answer is q/2\(\epsilon_0\).