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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

Expert verified
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

01

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.
02

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
03

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)
04

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
05

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.
06

Substitute r into D expression

For r=0.1m, substitute into the expression for D: D = 0.205a_r / (0.1m)² μC/m²
07

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)
08

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
09

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.
10

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
11

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Volume charge density is located as follows: \(\rho_{v}=0\) for \(\rho<1 \mathrm{~mm}\) and for \(\rho>2 \mathrm{~mm}, \rho_{v}=4 \rho \mu \mathrm{C} / \mathrm{m}^{3}\) for \(1<\rho<2 \mathrm{~mm} .\) (a) Calculate the total charge in the region \(0<\rho<\rho_{1}, 0

In a region in free space, electric flux density is found to be $$ \mathbf{D}=\left\\{\begin{array}{lr} \rho_{0}(z+2 d) \mathbf{a}_{z} \mathrm{C} / \mathrm{m}^{2} & (-2 d \leq z \leq 0) \\ -\rho_{0}(z-2 d) \mathbf{a}_{z} \mathrm{C} / \mathrm{m}^{2} & (0 \leq z \leq 2 d) \end{array}\right. $$ Everywhere else, \(\mathbf{D}=0 .\left(\right.\) a) Using \(\nabla \cdot \mathbf{D}=\rho_{v}\), find the volume charge density as a function of position everywhere. (b) Determine the electric flux that passes through the surface defined by \(z=0,-a \leq x \leq a,-b \leq y \leq b\). (c) Determine the total charge contained within the region \(-a \leq x \leq a\), \(-b \leq y \leq b,-d \leq z \leq d .(d)\) Determine the total charge contained within the region \(-a \leq x \leq a,-b \leq y \leq b, 0 \leq z \leq 2 d\).

A radial electric field distribution in free space is given in spherical coordinates as: $$ \begin{array}{l} \mathbf{E}_{1}=\frac{r \rho_{0}}{3 \epsilon_{0}} \mathbf{a}_{r} \quad(r \leq a) \\ \mathbf{E}_{2}=\frac{\left(2 a^{3}-r^{3}\right) \rho_{0}}{3 \epsilon_{0} r^{2}} \mathbf{a}_{r} \quad(a \leq r \leq b) \\ \mathbf{E}_{3}=\frac{\left(2 a^{3}-b^{3}\right) \rho_{0}}{3 \epsilon_{0} r^{2}} \mathbf{a}_{r} \quad(r \geq b) \end{array} $$ where \(\rho_{0}, a\), and \(b\) are constants. \((a)\) Determine the volume charge density in the entire region \((0 \leq r \leq \infty)\) by the appropriate use of \(\nabla \cdot \mathbf{D}=\rho_{v} \cdot(b) \mathrm{In}\) terms of given parameters, find the total charge, \(Q\), within a sphere of radius \(r\) where \(r>b\).

A certain light-emitting diode (LED) is centered at the origin with its surface in the \(x y\) plane. At far distances, the LED appears as a point, but the glowing surface geometry produces a far-field radiation pattern that follows a raised cosine law: that is, the optical power (flux) density in watts \(/ \mathrm{m}^{2}\) is given in spherical coordinates by $$ \mathbf{P}_{d}=P_{0} \frac{\cos ^{2} \theta}{2 \pi r^{2}} \mathbf{a}_{r} \quad \text { watts } / \mathrm{m}^{2} $$ where \(\theta\) is the angle measured with respect to the direction that is normal to the LED surface (in this case, the \(z\) axis), and \(r\) is the radial distance from the origin at which the power is detected. \((a)\) In terms of \(P_{0}\), find the total power in watts emitted in the upper half-space by the LED; (b) Find the cone angle, \(\theta_{1}\), within which half the total power is radiated, that is, within the range \(0<\theta<\theta_{1} ;\) ( \(c\) ) An optical detector, having a \(1-\mathrm{mm}^{2}\) cross-sectional area, is positioned at \(r=1 \mathrm{~m}\) and at \(\theta=45^{\circ}\), such that it faces the \(\mathrm{LED}\). If one milliwatt is measured by the detector, what (to a very good estimate) is the value of \(P_{0}\) ?

A uniform volume charge density of \(80 \mu \mathrm{C} / \mathrm{m}^{3}\) is present throughout the region \(8 \mathrm{~mm}10 \mathrm{~mm}\), find \(D_{r}\) at \(r=20 \mathrm{~mm}\).

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