Question

(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

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

(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.

(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: \(\nabla \cdot \mathbf{D} = \frac{1}{r^2} \frac{\partial (r^2 k \cdot \frac{Q}{r^2})}{\partial r}\)

(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.

(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 \(Q = \rho_{v 0} \cdot \frac{4\pi}{3} a^3\) \(\rho_{v 0} = \frac{3Q}{4\pi a^3}\)

(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.