A cube with edge lengths 1.6 m contains positively charged particles. Consider a Gaussian surface of
exactly the same dimensions surrounding the cube. The electric field caused by the charges is normal
to all the faces of the Gaussian cube. If the electric field at each face of the cube has a magnitude of
750 N/C, determine the average volume charge density in the cube.
To determine the average volume charge density in the cube, we need to use Gauss's law. According to Gauss's law, the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.
Given that the electric field at each face of the cube has a magnitude of 750 N/C, we can use this information to calculate the electric flux through one face of the Gaussian cube.
The electric flux, Φ, through a closed surface is given by the formula:
Φ = E * A
where E is the electric field and A is the area of the surface.
Since the electric field is normal (perpendicular) to all the faces of the Gaussian cube, the flux through each face will be the same.
Let's calculate the electric flux through one face:
Given:
Electric field, E = 750 N/C
Area of one face, A = (1.6 m)^2 = 2.56 m^2
Φ = E * A
= 750 N/C * 2.56 m^2
= 1920 Nm^2/C
Now, since the electric flux is the same for all six faces of the Gaussian cube, the total electric flux through the Gaussian surface is:
Total flux (Φ_total) = 6 * Φ
= 6 * 1920 Nm^2/C
= 11520 Nm^2/C
The total flux Φ_total is equal to the charge enclosed (Q_enc) divided by the permittivity of the medium (ε):
Φ_total = Q_enc / ε
Since the electric field is caused by the positively charged particles within the cube, the charge enclosed is the volume charge density (ρ) multiplied by the volume (V) of the cube:
Q_enc = ρ * V
Substituting this value into the equation, we get:
11520 Nm^2/C = (ρ * V) / ε
But we know that the volume (V) of the cube is (1.6 m)^3 = 4.096 m^3.
Therefore, the equation becomes:
11520 Nm^2/C = (ρ * 4.096 m^3) / ε
Now, we need to determine the permittivity (ε) of the medium. Since the problem does not provide this information, we can assume it to be the permittivity of vacuum, which is ε₀ = 8.854 x 10^-12 C^2/Nm^2.
Substituting the value of ε and rearranging the equation to solve for the volume charge density (ρ), we have:
ρ = (11520 Nm^2/C * ε₀) / 4.096 m^3
= (11520 Nm^2/C * 8.854 x 10^-12 C^2/Nm^2) / 4.096 m^3
Calculating this expression will give us the average volume charge density (ρ) in coulombs per cubic meter (C/m^3).
To determine the average volume charge density in the cube, we can first calculate the electric flux through one face of the Gaussian cube, and then use Gauss' Law to relate the flux to the enclosed charge.
1. Calculate the electric flux through one face of the Gaussian cube:
The electric field on each face of the cube has a magnitude of 750 N/C. We know that the electric flux (Φ) through a closed surface is given by the product of the electric field strength (E) and the area (A) of the surface. In this case, since the Gaussian cube has faces of dimensions 1.6 m, the area of each face is (1.6 m)² = 2.56 m².
Flux through one face = Electric field strength * Area of the face
Flux through one face = 750 N/C * 2.56 m²
Flux through one face = 1920 N m²/C (or equivalently, 1920 N m²/Coulomb)
2. Apply Gauss' Law to relate the flux to the enclosed charge:
Gauss' Law states that the electric flux through any closed surface is equal to the total enclosed charge divided by the permittivity of free space (ε₀), which is a constant value.
Flux through all the faces of the Gaussian cube = Electric flux through one face
From Gauss' Law, we have:
Flux through all the faces of the Gaussian cube = Total enclosed charge / ε₀
Since the Gaussian surface is exactly the same dimensions as the cube, the total enclosed charge is the charge contained within the cube.
3. Calculate the average volume charge density in the cube:
The average volume charge density (ρ) is defined as the total charge (Q) divided by the volume (V) of the cube.
ρ = Q / V
From step 2, we know that:
Flux through all the faces of the Gaussian cube = Total enclosed charge / ε₀
Comparing this with the definition of electric flux, E * A = Q / ε₀, where Q is the total enclosed charge, and A is the area of one face of the Gaussian cube.
So we can rewrite the equation as follows:
Q / ε₀ = 750 N/C * (2.56 m²) * 6 (since there are 6 faces in a cube)
Since the dimensions of the cube are not specified, we need to consider the volume of the cube. Let's assume the volume of the cube (V) is given by V = L³, where L is the length of an edge of the cube.
Comparing this with the definition of average volume charge density, ρ = Q / V, we can rewrite the equation as follows:
ρ = (750 N/C * (2.56 m²) * 6) / L³
Substituting the given edge length of the cube as 1.6 m, we can calculate the average volume charge density (ρ) as follows:
ρ = (750 N/C * (2.56 m²) * 6) / (1.6 m)³
ρ = (750 N/C * 15.36 m²) / (4.096 m³)
ρ ≈ 2824.22 C/m³
Therefore, the average volume charge density in the cube is approximately 2824.22 C/m³.