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Chapter 3: Problem 9

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 \(r=20 \mathrm{~mm}\).

Short Answer

Expert verified
Answer: The radial component of the electric displacement field is \(D_r(10 \text{ mm}) \approx 7.85 \text{ N/C}\) and \(D_r(20 \text{ mm}) \approx 1.96 \text{ N/C}\).

Step by step solution

01

Calculate total charge inside the sphere

To find the total charge inside the spherical surface with radius \(r=10\) mm, we need to integrate the volume charge density \(\rho_v\) over the given region (between radius \(r=8\) mm and \(r=10\) mm). Let \(Q\) be the total charge, \(r_1 = 8\) mm, \(r_2 = 10\) mm and \(\rho_v = 80 \times 10^{-6} \text{ C/m}^3\). The total charge \(Q\) can be found as: $$Q = \int_{V} \rho_v dV$$ Where \(dV\) is the volume element, given by \(dV = r^2 \sin \theta dr d\theta d\phi\) in spherical coordinates, with \(\theta\) and \(\phi\) representing the polar and azimuthal angles, respectively: $$Q = \int_{r_1}^{r_2} \int_{0}^{\pi} \int_{0}^{2\pi} (80 \times 10^{-6})r^2 \sin\theta dr d\theta d\phi$$
02

Perform the integration

Now, we will perform the integration: $$Q = \int_{r_1}^{r_2} (80 \times 10^{-6}) r^2 dr \cdot \int_{0}^{\pi}\sin\theta d\theta \cdot \int_{0}^{2\pi} d\phi$$ Solving each integral separately, we get: $$Q = (80 \times 10^{-6}) \left[\frac{r^3}{3}\right]_{r_1}^{r_2} \cdot [-\cos\theta]_{0}^{\pi} \cdot [\phi]_{0}^{2\pi}$$ $$Q = (80 \times 10^{-6}) \cdot \frac{(10^3 - 8^3)}{3} \cdot 2 \cdot 2\pi$$ Calculate the total charge: $$Q = 80 \times 10^{-6} \cdot \frac{936}{3} \cdot 4\pi \approx 3.14 \times 10^{-3} \text{ C}$$
03

Find \(D_r\) at \(r=10\) mm

To find the radial component of the electric displacement, \(D_r\), at \(r=10\) mm, we will use Gauss's law: $$\oint_{\Sigma} \vec{D} \cdot d\vec{\Sigma} = Q_{enclosed}$$ Since the problem has spherical symmetry, we can assume that the radial component of \(D_r\) is constant over the Gaussian surface. Thus, $$D_r \cdot \oint_{\Sigma} d\Sigma = Q_{enclosed}$$ Considering the Gaussian surface to be a sphere of radius \(r=10\) mm, the total surface area \(\Sigma\) is given by \(4\pi r^2\): $$D_r \cdot 4\pi (10 \times 10^{-3})^2 = 3.14 \times 10^{-3}$$ Solving for \(D_r\), $$D_r = \frac{3.14 \times 10^{-3}}{4\pi (10 \times 10^{-3})^2} = 7.85 \text{ N/C}$$
04

Find \(D_r\) at \(r=20\) mm

Now, let's find \(D_r\) at \(r=20\) mm, considering no charges for \(r>10\) mm. According to Gauss's law, $$D_r \cdot \oint_{\Sigma} d\Sigma = Q_{enclosed}$$ As the charge enclosed does not change, even though we have now doubled the radius, the enclosed charge is still \(Q = 3.14 \times 10^{-3}\) C. $$D_r \cdot 4\pi (20 \times 10^{-3})^2 = 3.14 \times 10^{-3}$$ Solving for \(D_r\) at \(r=20\) mm, $$D_r = \frac{3.14 \times 10^{-3}}{4\pi (20 \times 10^{-3})^2} = 1.96 \text{ N/C}$$ So, the radial component of the electric displacement field \(D_r\) is: (a) \(Q \approx 3.14 \times 10^{-3}\) C; (b) \(D_r(10 \text{ mm}) \approx 7.85 \text{ N/C}\); and (c) \(D_r(20 \text{ mm}) \approx 1.96 \text{ N/C}\).

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Most popular questions from this chapter

Spherical surfaces at \(r=2,4\), and \(6 \mathrm{~m}\) carry uniform surface charge densities of \(20 \mathrm{nC} / \mathrm{m}^{2},-4 \mathrm{n} \mathrm{C} / \mathrm{m}^{2}\), and \(\rho_{\mathrm{so}}\), respectively. \((a)\) Find \(\mathbf{D}\) at \(r=1\), 3 , and \(5 \mathrm{~m}\). (b) Determine \(\rho_{S 0}\) such that \(\mathbf{D}=0\) at \(r=7 \mathrm{~m}\).

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