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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.

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, 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.