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: $$ Q = 4 \pi \rho_0 \int_0^a r^2 dr - 4 \pi \rho_0 \int_a^b r^3 dr $$ $$ Q = 4 \pi \rho_0 \left[\frac{1}{3} r^3 \Big|_0^a - \frac{1}{4} r^4 \Big|_a^b \right] $$ $$ Q = 4 \pi \rho_0 \left[\frac{1}{3} a^3 - \frac{1}{4} (b^4 - a^4) \right] $$

Key concepts

Gaussian Surface

A Gaussian surface is an imaginary closed surface used in Gauss's Law to calculate electric fields or electric flux through a surface. Gauss's Law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of space. In mathematical terms, it's expressed as: \[ \Phi_E = \oint \mathbf{E} \cdot \mathbf{a} \, dA = \frac{Q_{enc}}{\epsilon_0} \] In spherical coordinates, Gaussian surfaces become particularly useful when dealing with symmetrical charge distributions. By choosing a sphere as the Gaussian surface, the symmetry simplifies calculations. This allows us to treat the electric field as constant over the surface, making integrals straightforward. Using a Gaussian surface can significantly simplify finding the electric field due to a given charge distribution. It leverages symmetry to make otherwise complex integrals manageable. In problems where symmetry exists, applying Gaussian surfaces can save a lot of time and aid in a deeper understanding of the concept of electric flux in relation to charge distributions.

Volume Charge Density

Volume charge density, commonly denoted as \(\rho_v\), indicates how much electric charge is distributed within a given volume. It relates to Gauss's Law through the divergence of electric displacement, \(abla \cdot \mathbf{D}\), which equals the volume charge density \(\rho_v\). In mathematical form, Gauss's Law for continuous charge distribution states: \[ abla \cdot \mathbf{D} = \rho_v \] For spherical systems, it is crucial to apply this law in spherical coordinates, where symmetry can simplify calculations. As seen in problems with spherical charges, the volume charge density determines how charges are distributed across different regions (e.g., inside a sphere or shell). This makes it easier to evaluate how charges influence the electric field within those regions. Understanding volume charge density helps in comprehensively analyzing various electrostatic scenarios. With the information about \(\rho_v\), one can predict electric behavior and calculate fields and potentials across space in practical scenarios as well.

Electric Displacement

Electric displacement, denoted \(\mathbf{D}\), is an auxiliary electric field that quantifies free and bound charge separation in dielectric materials. It's directly related to the electric field \(\mathbf{E}\) and the permittivity of free space \(\epsilon_0\) in vacuum or air environments. The relationship is given by: \[ \mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P} \] where \(\mathbf{P}\) is the polarization of the medium. In free space, this simplifies to \(\mathbf{D} = \epsilon_0 \mathbf{E}\). In our problem, understanding \(\mathbf{D}\) is necessary because it helps in using Gauss's Law efficiently. It allows for the calculation of electric fields when charges are distributed across different mediums. For spherical distributions, knowing \(\mathbf{D}\) aids in determining the flux through a surface, which correlates to the charge distribution inside. Mastering the concept of electric displacement improves insight into how dielectric materials react to electric fields, making it fundamental for both theoretical and applied electromagnetism.

Spherical Coordinates

Spherical coordinates are vital when dealing with spherical symmetry in electric fields and charge distributions. In these coordinates, the position of a point in space is given in terms of radial distance \(r\), polar angle \(\theta\), and azimuthal angle \(\phi\). This system is particularly beneficial in problems involving radially symmetric fields, like those emanating from point charges or spherical charge distributions. It simplifies the use of Gauss's Law due to the symmetry, which allows for easier integration and evaluation of field quantities. When using spherical coordinates, the electric field \(\mathbf{E}\) can often be expressed as a function of the radial unit vector \(\mathbf{a}_r\), primarily depending on the radial distance \(r\). It's common to see field expressions defined over different regions, which can be handled effectively using spherical coordinates. Embracing spherical coordinates is crucial for tackling many physics and engineering problems, promoting a more straightforward approach to analyzing the interaction between electric fields and spherical charge distributions.