A charge of + 5C is placed at one corner of side m , What is the flux through the front face and back face ?

To determine the electric flux through the front and back faces of a cube when a charge of +5C is placed at one corner, we can apply Gauss's Law. This law states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. Let's break this down step by step.

Understanding Electric Flux

Electric flux (Φ) is defined as the product of the electric field (E) and the area (A) through which it passes, and it can be mathematically expressed as:

Φ = E × A × cos(θ)

Here, θ is the angle between the electric field lines and the normal (perpendicular) to the surface. When the field is perpendicular to the surface, cos(θ) equals 1, simplifying our calculations.

Applying Gauss's Law

According to Gauss's Law, the total electric flux through a closed surface is given by:

Φ_total = Q_enclosed / ε₀

Where:

• Φ_total is the total electric flux through the closed surface.
• Q_enclosed is the total charge enclosed within the surface.
• ε₀ is the permittivity of free space, approximately equal to 8.85 × 10⁻¹² C²/(N·m²).

Calculating the Flux Through the Cube

In this scenario, the charge of +5C is located at one corner of the cube. Since the charge is at a corner, it is important to note that it is not fully enclosed by the cube. In fact, only one-eighth of the charge contributes to the flux through the cube. This is because the cube shares this corner with seven other cubes in a three-dimensional lattice.

Thus, the enclosed charge for our cube is:

Q_enclosed = +5C / 8 = +0.625C

Calculating Total Flux

Now, we can calculate the total electric flux through the entire surface of the cube using Gauss's Law:

Φ_total = Q_enclosed / ε₀ = 0.625C / (8.85 × 10⁻¹² C²/(N·m²))

Calculating this gives:

Φ_total ≈ 7.06 × 10⁹ N·m²/C

Distributing Flux Through the Faces

The total flux calculated is distributed equally among the six faces of the cube. Therefore, the flux through each face of the cube is:

Φ_face = Φ_total / 6 ≈ 7.06 × 10⁹ N·m²/C / 6 ≈ 1.18 × 10⁹ N·m²/C

Specific Faces: Front and Back

Since the charge is at a corner, the front and back faces of the cube will each receive a portion of the total flux. However, due to symmetry and the distribution of the electric field, the flux through the front face and the back face will be equal:

Φ_front = Φ_back ≈ 1.18 × 10⁹ N·m²/C

In summary, the electric flux through both the front and back faces of the cube is approximately 1.18 × 10⁹ N·m²/C. This example illustrates the application of Gauss's Law in a practical scenario, highlighting how charges influence electric fields and flux in three-dimensional space.