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.