Question

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 \(r>b\).

Short answer

Answer: The expression for the total charge Q within a sphere of radius r where r > b is given by: $$ Q = 4 \pi \rho_0 \left[\frac{1}{3} a^3 - \frac{1}{4} (b^4 - a^4) \right] $$ where \(\rho_0\) is the volume charge density in the given regions, \(a\) is the inner radius of region 2, and \(b\) is the outer radius of region 2.

Step-by-step solution

Remember Gauss's Law in spherical coordinates

In spherical coordinates, the Gauss's Law is given by: $$ \nabla \cdot \mathbf{D} = \frac{1}{r^2} \frac{\partial{(r^2D_r)}}{\partial{r}} = \rho_v $$ where \(D_r\) is the radial component of the electric displacement.

Relate electric displacement with electric field

The electric displacement \(\mathbf{D}\) is related to the electric field \(\mathbf{E}\) by the equation: $$ \mathbf{D} = \epsilon_0 \mathbf{E} $$

Calculate volume charge density for each region

For each region, we will use the provided electric field and Gauss's law to calculate the volume charge density. Region 1: \(r \leq a\) $$ \rho_{v1} = \frac{1}{r^2} \frac{\partial{(r^2(\epsilon_0 E_{1r}))}}{\partial{r}} = \frac{1}{r^2} \frac{\partial{(r^3 \frac{r \rho_0}{3 \epsilon_0}))}{\partial{r}} = \rho_0 $$ Region 2: \(a \leq r \leq b\) $$ \rho_{v2} = \frac{1}{r^2} \frac{\partial{(r^2(\epsilon_0 E_{2r}))}}{\partial{r}} = \frac{1}{r^2} \frac{\partial{(2 a^{3} \rho_{0} - r^3 \rho_{0})}{\partial{r}} = -\frac{3\rho_0 r^3}{3 \epsilon_0} \cdot \frac{1}{r^2} = -\rho_0 r $$ Region 3: \(r \geq b\) $$ \rho_{v3} = \frac{1}{r^2} \frac{\partial{(r^2(\epsilon_0 E_{3r}))}}{\partial{r}} = 0 $$

Calculate the total charge Q within a sphere of radius r where r>b

For calculating total charge Q, we will integrate the volume charge density over the sphere of radius r: $$ Q = \int_0^a \rho_{v1} 4 \pi r^2 dr + \int_a^b \rho_{v2} 4 \pi r^2 dr + \int_b^r \rho_{v3} 4 \pi r^2 dr $$ Since \(\rho_{v3} = 0\), the last integral will be 0. $$ Q = \int_0^a \rho_{0} 4 \pi r^2 dr + \int_a^b -\rho_{0}r 4 \pi r^2 dr $$ Now, evaluate the remaining integrals: $$ Q = 4 \pi \rho_0 \int_0^a r^2 dr - 4 \pi \rho_0 \int_a^b r^3 dr $$ $$ Q = 4 \pi \rho_0 \left[\frac{1}{3} r^3 \Big|_0^a - \frac{1}{4} r^4 \Big|_a^b \right] $$ $$ Q = 4 \pi \rho_0 \left[\frac{1}{3} a^3 - \frac{1}{4} (b^4 - a^4) \right] $$