Question

Let \(\mathbf{D}=5.00 r^{2} \mathbf{a}_{r} \mathrm{mC} / \mathrm{m}^{2}\) for \(r \leq 0.08 \mathrm{~m}\) and \(\mathbf{D}=0.205 \mathrm{a}_{r} / r^{2} \mu \mathrm{C} / \mathrm{m}^{2}\) for \(r \geq 0.08 \mathrm{~m} .(a)\) Find \(\rho_{v}\) for \(r=0.06 \mathrm{~m} .(b)\) Find \(\rho_{v}\) for \(r=0.1 \mathrm{~m} .(c)\) What surface charge density could be located at \(r=0.08 \mathrm{~m}\) to cause \(\mathbf{D}=0\) for \(r>0.08 \mathrm{~m} ?\)

Short answer

Answer: The volume charge density (ρ_v) at r=0.06m is 5.00 * (0.06m)² * a_r mC/m² and at r=0.1m is 0.205a_r / (0.1m)² μC/m². The surface charge density (σ) at r=0.08m for D=0 is -0.205a_r / (0.08m)² μC/m².

Step-by-step solution

a) Find ρ_v for r=0.06m:

First, we have to use the expression for D for r ≤ 0.08m, which is given as D = 5.00r²a_r mC/m² for r ≤ 0.08m. Next, we will substitute r=0.06m into the equation. Using the formula D = ε₀E, we will find the electric field E and then determine the volume charge density ρ_v.

Substitute r into D expression

For r=0.06m, substitute into the expression for D: D = 5.00 * (0.06m)² * a_r mC/m²

Find E using D = ε₀E formula

ε₀ = 8.854 x 10^{-12} C²/Nm² (vacuum permittivity constant). Now, solve for E: E = D/ ε₀ = (5.00 * (0.06m)² * a_r mC/m²) / (8.854 x 10^{-12} C²/Nm²)

Find ρ_v using E = ρ_v / ε₀ formula

Now, we know that E = ρ_v / ε₀. Substituting E into this formula, we find: ρ_v = ε₀E = 5.00 * (0.06m)² * a_r mC/m²

b) Find ρ_v for r=0.1m:

Now, we have to use the expression for D for r ≥ 0.08m, which is given as D = 0.205a_r / r² μC/m². Next, we will substitute r=0.1m into the equation. Using the D = ε₀E formula, we will find the electric field E and then determine the volume charge density ρ_v.

Substitute r into D expression

For r=0.1m, substitute into the expression for D: D = 0.205a_r / (0.1m)² μC/m²

Find E using D = ε₀E formula

Again, ε₀ = 8.854 x 10^{-12} C²/Nm². Now, solve for E: E = D/ ε₀ = (0.205a_r / (0.1m)² μC/m²) / (8.854 x 10^{-12} C²/Nm²)

Find ρ_v using E = ρ_v / ε₀ formula

Substitute E into the formula E = ρ_v / ε₀, and we find: ρ_v = ε₀E = 0.205a_r / (0.1m)² μC/m²

c) Surface charge density at r=0.08m for D=0:

We need to find the surface charge density (σ) that would make D=0 for r≥0.08m. Since we already have the expression for D for r≥0.08m, we simply need to set D=0 and solve for σ.

Set D=0 for r≥0.08m

For r=0.08m, D expression is: D = 0.205a_r / r² μC/m² Set D = 0 and substitute r=0.08m: 0 = 0.205a_r / (0.08m)² μC/m²

Solve for σ

Now, we can solve the equation for σ: σ = -0.205a_r / (0.08m)² μC/m² Thus, the surface charge density at r=0.08m for D=0 is -0.205a_r / (0.08m)² μC/m².

Key concepts

Volume Charge Density

Volume charge density, denoted by \( \rho_v \), is a measure of the amount of electric charge per unit volume in a region of space. When it comes to calculating \( \rho_v \) from the dielectric displacement field \( \mathbf{D} \), we typically use Gauss's law in differential form. This approach relates \( \mathbf{D} \) to \( \rho_v \) in materials through the equation: \( abla \cdot \mathbf{D} = \rho_v \). This relationship helps us decide the charge distribution within a given space.
To determine \( \rho_v \) for \( r = 0.06 \, \text{m} \), we begin by using the provided expression for \( \mathbf{D} \) in the region \( r \leq 0.08 \, \text{m} \). The expression is \( \mathbf{D} = 5.00r^2 \mathbf{a}_r \, \text{mC/m}^2 \). By substituting \( r = 0.06 \, \text{m} \) into this equation, we can calculate \( D \). To find the electric field \( \mathbf{E} \), we use the relationship \( \mathbf{D} = \varepsilon_0 \mathbf{E} \), where \( \varepsilon_0 \) is the permittivity of free space. Solving for \( \mathbf{E} \) then lets us determine \( \rho_v = \varepsilon_0 E \). This calculation process gives us an understanding of how charge is distributed in the region at specific distances.

Electric Field Calculation

The electric field \( \mathbf{E} \) is a vector field around charged particles. It represents the force a charge experiences per unit charge. Calculating \( \mathbf{E} \) involves using the dielectric displacement field \( \mathbf{D} \), which simplifies in scenarios involving medium permittivity. We use the formula \( \mathbf{D} = \varepsilon_0 \mathbf{E} \), rearranging it to find \( \mathbf{E} = \mathbf{D} / \varepsilon_0 \).
For example, finding \( \mathbf{E} \) at \( r = 0.1 \, \text{m} \) involves using the \( \mathbf{D} \) expression given for \( r \geq 0.08 \, \text{m} \). We substitute \( r = 0.1 \, \text{m} \) into the equation \( \mathbf{D} = 0.205 \mathbf{a}_r / r^2 \, \mu \text{C/m}^2 \) to solve for \( \mathbf{D} \).
Dividing this \( \mathbf{D} \) by \( \varepsilon_0 \) (\( 8.854 \times 10^{-12} \text{ C}^2/\text{Nm}^2 \)) will provide us \( \mathbf{E} \). This process illustrates how knowing the displacement field helps us determine the electric influences in different points inside the material.

Surface Charge Density

Surface charge density, represented by \( \sigma \), quantifies the charge per unit area on a surface. It's crucial when distinguishing between charge distributions on boundaries rather than throughout volumes. In our scenario, the challenge is to calculate \( \sigma \) at \( r = 0.08 \, \text{m} \) to make the dielectric displacement field \( \mathbf{D} \) zero for \( r > 0.08 \, \text{m} \). This means ensuring the boundary at \( r = 0.08 \, \text{m} \) contains enough negative charge to stop \( \mathbf{D} \) from propagating further.
To find \( \sigma \), we use the condition at \( r = 0.08 \, \text{m} \), setting \( \mathbf{D} = 0.205 \mathbf{a}_r / (r)^2 \mu \text{C/m}^2 \). Solving \( \sigma \) gives us insight into this charge's configuration.
By setting \( \mathbf{D} = 0 \) for \( r = 0.08 \, \text{m} \), we derive \( \sigma = -0.205 \mathbf{a}_r / (0.08)^2 \mu \text{C/m}^2 \). Understanding \( \sigma \) clarifies how surface charges must be adjusted to influence the electric field's behavior across boundaries.