Question

Volume charge density is located in free space as \(\rho_{v}=2 e^{-1000 r} \mathrm{nC} / \mathrm{m}^{3}\) for \(0

Short answer

Answer: The total charge enclosed by the spherical surface of radius 1mm is approximately 7.84 pC, and the electric flux density on the surface is approximately 140 N/C.

Step-by-step solution

Calculate Total Charge

To find the total charge enclosed by a sphere of radius r = 1mm, let's first define the given volume charge density: Ρv(r) = 2e^{-1000r} nC/m³ (for 0 < r < 1mm) Now, we need to integrate this charge density over the whole volume of the sphere: Total charge (Q) = ∫∫∫Ρv(r) dv Where the limits of integration are from r = 0 to r = 1mm, and we need to convert the radius into meters (0 to 0.001m). And dv = r²sinθ dr dθ dϕ where θ is from 0 to π, and ϕ is from 0 to 2π. Q = ∫(0 to 0.001) ∫(0 to π) ∫(0 to 2π) 2e^{-1000r} r² sinθ dr dθ dϕ

Perform the Integration

Let's solve the volume integral one variable at a time. First, we perform the integral with respect to ϕ: Q = ∫(0 to 0.001) ∫(0 to π) [2π * 2e^{-1000r} r² sinθ] dr dθ Now, we perform the integral with respect to θ: Q = ∫(0 to 0.001) [2π * -2e^{-1000r} r² (cosθ from 0 to π)] dr Q = ∫(0 to 0.001) [4π * 2e^{-1000r} r²] dr Finally, we perform the integral with respect to r: Q = -4π * 2 * 10^{-6} [e^{-1000r} r²/(-1000) + 2e^{-1000r} r/(-1000)^2 + 2e^{-1000r}/(-1000)^3] from r=0 to r=(1mm = 0.001m) By plugging r = 0.001m and r = 0 into the above expression and calculating we get, Q ≈ 7.84 * 10^{-12} C The total charge enclosed by the sphere is approximately 7.84 pC.

Calculate Electric Flux Density using Gauss's Law

Now that we have the total enclosed charge, we can use Gauss's Law to find the electric flux density at r=1mm. Gauss's Law states that the total electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space (ε0). Flux = Q / ε0 The electric flux density on the surface (Dr) is equal to the total electric flux divided by the surface area of the sphere (A). Surface area of the sphere A = 4πr². In this case, r = 1mm = 0.001m. Dr = Flux / A = (Q / ε0) / (4πr²) By plugging the known values into the equation: ε0 = 8.85 * 10^{-12} C²/Nm² Q ≈ 7.84 * 10^{-12} C r = 0.001m Dr ≈ ((7.84 * 10^{-12} C) / (8.85 * 10^{-12} C²/Nm²)) / (4π(0.001m)²) Dr ≈ 140 N/C The electric flux density on the surface r = 1mm is approximately 140 N/C.

Key concepts

Volume Charge Density

Volume charge density, denoted as \( \rho_v \), represents how much electric charge is present in a given volume of space. It's often measured in coulombs per cubic meter (\( \text{C/m}^3 \)). In our exercise, the volume charge density is given by \( \rho_v = 2e^{-1000r} \text{ nC/m}^3 \) for \(0 < r < 1\text{ mm}\), and \( \rho_v = 0 \) elsewhere. This equation tells us that the charge density decreases exponentially as the distance \( r \) increases, and only exists within a spherical region of radius 1 millimeter. The value of \( \rho_v \) becomes very small as \( r \) approaches 1 mm, and it drops to zero outside this sphere. This implies that the charges are mainly concentrated near the center of the sphere, diminishing as we move outward.

Electric Flux Density

Electric flux density, also known as the displacement field, is represented by the symbol \( \mathbf{D} \). It relates how an electric field influences the distribution of electric charges in a material. It's expressed in units of coulombs per square meter (\( \text{C/m}^2 \)). In relation to Gauss's Law, the total outward electric flux through a closed surface is proportional to the charge enclosed by that surface.In our scenario, after determining the total charge within the sphere using the volume charge density, we calculate the electric flux density at \( r = 1 \text{ mm} \) using Gauss's Law. The equation used here is \( \mathbf{D} = \frac{Q}{\varepsilon_0} \), where \( Q \) is the total enclosed charge and \( \varepsilon_0 \) is the permittivity of free space. This value gives us an insight into how crowded the electric lines of flux are on the spherical surface at this specific radius.

Spherical Surface Integration

Spherical surface integration is a method used to calculate quantities over a spherical region. It is crucial when the problem involves a sphere or a circular symmetry, such as the given charge density. For our exercise, spherical coordinates \( (r, \theta, \phi) \) are used, where \( r \) is the radial distance, \( \theta \) is the polar angle, and \( \phi \) is the azimuthal angle.The differential volume element in spherical coordinates is \( dv = r^2 \sin(\theta) \, dr \, d\theta \, d\phi \). For integration:- \( r \) ranges from 0 to 0.001 meters (1 mm),- \( \theta \) ranges from 0 to \( \, \pi \),- \( \phi \) ranges from 0 to \(2 \pi \).These integrals allow us to consider all points in the sphere when calculating the total charge, ensuring that we correctly account for how the charge density decreases with increasing \( r \).

Permittivity of Free Space

The permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental physical constant that describes how electric fields interact with materials in a vacuum. It is crucial in the calculations of electric fields and flux densities, especially when using Gauss's Law.The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \), highlighting that it is a very small number. This means that in a vacuum, the ability of space to permit electric field lines is very limited.In our exercise, this constant is used to relate the total electric flux to the enclosed charge \( Q \) through the equation \( \text{Flux} = \frac{Q}{\varepsilon_0} \). Understanding \( \varepsilon_0 \) helps us grasp how quickly or slowly an electric field can expand in a vacuum, and how charge distributions can affect the strength and reach of the field.