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Chapter 3: Problem 23

(a) A point charge \(Q\) lies at the origin. Show that div \(\mathbf{D}\) is zero everywhere except at the origin. (b) Replace the point charge with a uniform volume charge density \(\rho_{v 0}\) for \(0

Short Answer

Expert verified
Answer: For a point charge Q at the origin, the divergence of the electric flux density field is zero everywhere except at the origin. For a uniform volume charge density ρ_v0 for 0<r<a, the divergence of the electric flux density field is \(\frac{3Q}{4\pi a^3}\) for 0<r<a, and zero for r≥a.

Step by step solution

01

(a) Given Information and Setting up the Problem

We are given a point charge Q at the origin. Given this setup, the electric flux density \(\mathbf{D}\) for a point charge can be given by: \(\mathbf{D} = k \cdot \frac{Q}{r^2} \hat{r}\), where \(\mathbf{D}\) is the electric flux density vector, \(k\) is the Coulomb's constant, \(Q\) is the charge, \(r\) is the distance from the origin, and \(\hat{r}\) is the radial unit vector. We use Gauss's Law (\(\nabla \cdot \mathbf{D} = \rho_v\)) to find the divergence of \(\mathbf{D}\), where \(\rho_v\) is the volume charge density.
02

(a) Find the Divergence of \(\mathbf{D}\)

Calculating the divergence of the electric flux density vector using spherical coordinates: \(\nabla \cdot \mathbf{D} = \frac{1}{r^2} \frac{\partial (r^2 D_r)}{\partial r} + \frac{1}{r\sin{\theta}}\frac{\partial(D_\theta \sin{\theta})}{\partial \theta} + \frac{1}{r\sin{\theta}}\frac{\partial D_\phi}{\partial \phi}\) Since \(\mathbf{D} = k \cdot \frac{Q}{r^2} \hat{r}\), \(D_\theta=0\) and \(D_\phi=0\). Substitute into our divergence equation: \(\nabla \cdot \mathbf{D} = \frac{1}{r^2} \frac{\partial (r^2 k \cdot \frac{Q}{r^2})}{\partial r}\)
03

(a) Evaluate the Derivative and Result

Evaluate the derivative with respect to r: \(\nabla \cdot \mathbf{D} = \frac{1}{r^2} \frac{\partial (kQ)}{\partial r} = 0\) Thus, the divergence of the electric flux density is zero everywhere except at the origin. At the origin, we cannot evaluate the divergence because the electric flux density vector is undefined.
04

(b) Replace Point Charge with Uniform Volume Charge Density

We are given a uniform volume charge density \(\rho_{v 0}\) for \(0 < r < a\), such that the total charge is the same as that of the point charge in part (a). To relate \(\rho_{v 0}\), Q, and a, use: total charge = volume charge density × volume \(Q = \rho_{v 0} \cdot \frac{4\pi}{3} a^3\) \(\rho_{v 0} = \frac{3Q}{4\pi a^3}\)
05

(b) Find the Divergence of \(\mathbf{D}\) for Uniform Volume Charge Density

Using Gauss's Law: \(\nabla \cdot \mathbf{D} = \rho_{v 0}\) Substitute the expression for \(\rho_{v 0}\): \(\nabla \cdot \mathbf{D} = \frac{3Q}{4\pi a^3}\) Therefore, for \(0 < r < a\), the divergence of the electric flux density is \(\frac{3Q}{4\pi a^3}\). Outside this region (i.e., \(r \geq a\)), the divergence of the electric flux density is zero.

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Most popular questions from this chapter

(a) A flux density field is given as \(\mathbf{F}_{1}=5 \mathbf{a}_{z} .\) Evaluate the outward flux of \(\mathbf{F}_{1}\) through the hemispherical surface, \(r=a, 0<\theta<\pi / 2,0<\phi<2 \pi\) (b) What simple observation would have saved a lot of work in part \(a ?\) (c) Now suppose the field is given by \(\mathbf{F}_{2}=5 z \mathbf{a}_{z} .\) Using the appropriate surface integrals, evaluate the net outward flux of \(\mathbf{F}_{2}\) through the closed surface consisting of the hemisphere of part \(a\) and its circular base in the \(x y\) plane. ( \(d\) ) Repeat part \(c\) by using the divergence theorem and an appropriate volume integral.

An electric field in free space is \(\mathbf{E}=\left(5 z^{2} / \epsilon_{0}\right) \hat{\mathbf{a}}_{z} \mathrm{~V} / \mathrm{m}\). Find the total charge contained within a cube, centered at the origin, of \(4-\mathrm{m}\) side length, in which all sides are parallel to coordinate axes (and therefore each side intersects an axis at \(\pm 2\) ).

A cube is defined by \(1

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 \(r=7 \mathrm{~m}\).

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 \(-a \leq x \leq a,-b \leq y \leq b, 0 \leq z \leq 2 d\).
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