Figure below shows a closed Gaussian surface in the shape of a cube of edge length l. It lies in a region where the electric field is given by = (ax + a) + b + c. What is the net charge contained by the cube in terms of the given variables? Use ε0 if necessary.

Question

Figure below shows a closed Gaussian surface in the shape of a cube of edge length l. It lies in a region where the electric field is given by = (ax + a) + b + c. What is the net charge contained by the cube in terms of the given variables? Use ε0 if necessary.

Answer 1

Without a figure, this can't be done. THe point is that the surface integral of E dot dS is = to the charge inside.

Answer 2

To find the net charge contained by the cube, we need to use Gauss's Law.

Gauss's Law states that the flux of the electric field through a closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space (ε₀).

Flux (Φ) through a closed surface is given by:
Φ = ∫∫ E · dA

In this case, our closed surface is in the shape of a cube. Let's assume that the electric field given is:
E = (ax + a) + b + c

To calculate the flux, we need to integrate the dot product of E with the area vector (dA) over the entire surface.

The flux through each face of the cube is given by:
Φ = ∫∫ E · dA = ∫∫ (E · n̂) dA

where n̂ is the unit normal vector to the surface and dA is the differential area element.

Since the cube has six faces, we need to calculate the flux through each face and sum them up to get the total flux through the closed surface.

Let's denote the sides of the cube as:
1. Front face: A₁
2. Back face: A₂
3. Right face: A₃
4. Left face: A₄
5. Top face: A₅
6. Bottom face: A₆

The unit normal vectors for each face are:
1. Front face: n̂₁ = -ẑ
2. Back face: n̂₂ = ẑ
3. Right face: n̂₃ = x̂
4. Left face: n̂₄ = -x̂
5. Top face: n̂₅ = ŷ
6. Bottom face: n̂₆ = -ŷ

The flux through each face can be calculated as follows:
1. Φ₁ = ∫∫ (E · n̂₁) dA = ∫∫ (-(az + a) · (-ẑ)) dA = ∫∫ (az + a) dA = (az + a) ∫∫ dA = (az + a) A₁
2. Φ₂ = ∫∫ (E · n̂₂) dA = ∫∫ ((az + a) · ẑ) dA = ∫∫ (az + a) dA = (az + a) ∫∫ dA = (az + a) A₂
3. Φ₃ = ∫∫ (E · n̂₃) dA = ∫∫ ((ax + a) · x̂) dA = ∫∫ (ax + a) dA = (ax + a) ∫∫ dA = (ax + a) A₃
4. Φ₄ = ∫∫ (E · n̂₄) dA = ∫∫ (-(ax + a) · (-x̂)) dA = ∫∫ (ax + a) dA = (ax + a) ∫∫ dA = (ax + a) A₄
5. Φ₅ = ∫∫ (E · n̂₅) dA = ∫∫ ((b) · ŷ) dA = ∫∫ (b) dA = b ∫∫ dA = b A₅
6. Φ₆ = ∫∫ (E · n̂₆) dA = ∫∫ (-(b) · (-ŷ)) dA = ∫∫ (b) dA = b ∫∫ dA = b A₆

Now, we can sum up the flux through all six faces to get the total flux (Φ) through the closed surface:
Φ = Φ₁ + Φ₂ + Φ₃ + Φ₄ + Φ₅ + Φ₆

Next, we equate the flux (Φ) to the net charge enclosed (Q) divided by ε₀:
Φ = Q / ε₀

Rearranging the equation, we can solve for the net charge (Q):
Q = Φ * ε₀

Therefore, to find the net charge contained by the cube, you need to calculate the total flux (Φ) using the expressions for Φ₁, Φ₂, Φ₃, Φ₄, Φ₅, Φ₆, and sum them up. Then, multiply the total flux by the permittivity of free space (ε₀) to find the net charge (Q) contained by the cube.