Problem 58 - University Physics with Modern Physics

Question

22.58. A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho(r)\) given as follows: $$ \begin{array}{ll}{\rho(r)=\rho_{0}(1-4 r / 3 R)} & {\text { for } r \leq R} \\\ {\rho(r)=0} & {\text { for } r \geq R}\end{array} $$ where \(\rho_{0}\) is a positive constant. (a) Find the total charge contained in the charge distribution. (b) Obtain an expression for the electric field in the region \(r \geq R .\) (c) Obtain an expression for the electric field in the region \(r \leq R .(d)\) Graph the electric-field magnitude \(E\) as a function of \(r .(e)\) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field. $$ \oint \overrightarrow{\boldsymbol{g}} \cdot d \overrightarrow{\boldsymbol{A}}=-4 \pi G m $$ (b) By following the same logical steps used in Section 22.3 to obtain Gauss's law for the electric field, show the flux of \(\overrightarrow{\boldsymbol{g}}\) through any closed surface is given by $$ \oint \overrightarrow{\boldsymbol{g}} \cdot d \boldsymbol{A}=-4 \pi G M_{\mathrm{encl}} $$

Step-by-Step Solution

Answer

(a) \( Q = \frac{4\pi \rho_0 R^3}{9} \). (b) \( E = \frac{\rho_0 R^3}{9 \varepsilon_0 r^2} \) for \( r \geq R \). (c) \( E = \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{r^2}{R} \right) \) for \( r \leq R \). (e) Max at \( r = \frac{R}{2} \), \( E_{\text{max}} = \frac{\rho_0 R}{12 \varepsilon_0} \).

Step 1: Calculate Total Charge Within Distribution

The total charge within the region \( r \leq R \) is given by the integral \( Q = \int \rho(r) dV \). The volume element in spherical coordinates is \( dV = 4\pi r^2 dr \). Thus, the charge can be calculated as follows:\[ Q = \int_0^R \rho_0 \left(1 - \frac{4r}{3R}\right) 4\pi r^2 dr \]Substitute \( \rho(r) \) into the integral:\[ Q = 4\pi \rho_0 \int_0^R \left(r^2 - \frac{4r^3}{3R}\right) dr \]Integrate:\[ Q = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{4r^4}{12R} \right]_0^R \]Calculating from \( 0 \) to \( R \):\[ Q = 4\pi \rho_0 \left[ \frac{R^3}{3} - \frac{R^3}{3} \right] = \frac{4\pi \rho_0 R^3}{9} \]

Step 2: Electric Field for r ≥ R (Outside the Sphere)

By symmetry and Gauss's Law, the electric field for \( r \geq R \) acts as if all the charge is concentrated at the center. Therefore:\[ E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \]Substituting \( Q = \frac{4\pi \rho_0 R^3}{9} \):\[ E = \frac{1}{\varepsilon_0} \cdot \frac{\frac{4\pi \rho_0 R^3}{9}}{4\pi r^2} = \frac{\rho_0 R^3}{9 \varepsilon_0 r^2} \]

Step 3: Electric Field for r ≤ R (Inside the Sphere)

For \( r \leq R \), the electric field is determined by the charge inside a Gaussian surface of radius \( r \).First, determine the enclosed charge:\[ q_{\text{enc}} = 4\pi \rho_0 \int_0^r \left( r'^2 - \frac{4r'^3}{3R} \right) dr' \]Integrate:\[ q_{\text{enc}} = 4\pi \rho_0 \left[ \frac{r'^3}{3} - \frac{r'^4}{3R} \right]_0^r = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right) \]Apply Gauss's Law:\[ E \cdot 4\pi r^2 = \frac{q_{\text{enc}}}{\varepsilon_0} \]\[ E = \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{r^2}{R} \right) \]

Step 4: Graph the Electric Field Magnitude E as a Function of r

The electric field magnitude \( E \) can be graphed as follows:- For \( r \leq R \), use \( E = \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{r^2}{R} \right) \).- For \( r \geq R \), use \( E = \frac{\rho_0 R^3}{9 \varepsilon_0 r^2} \).Plot \( E \) versus \( r \) highlighting the behavior for both regions, ensuring continuity at \( r = R \).

Step 5: Find Maximum Electric Field Value and Location

To find the maximum electric field value inside the sphere for \( r \leq R \), calculate the derivative of \( E(r) = \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{r^2}{R} \right) \) and set it to zero.Calculate:\[ \frac{dE}{dr} = \frac{\rho_0}{3 \varepsilon_0} \left( 1 - \frac{2r}{R} \right) \]Setting the derivative to zero gives:\[ 1 - \frac{2r}{R} = 0 \implies r = \frac{R}{2} \]Substitute \( r = \frac{R}{2} \) into \( E(r) \) to find the maximum field:\[ E_{\text{max}} = \frac{\rho_0 R}{12 \varepsilon_0} \]

Key Concepts

Charge Distribution

The distribution of charge describes how charge is spread across a given region. In this problem, we have a spherically symmetric charge distribution, which means the charge density only depends on the distance from the center, not the direction. The charge density \( \rho(r) \) is given by different formulas depending on the location \( r \). For \( r \leq R \), the density is \( \rho_0(1-\frac{4r}{3R}) \), and for \( r \geq R \), the density is zero, indicating all the charge is within a sphere of radius \( R \).
This type of problem requires integrating the charge density over a volume to find total charge:

• The integral's limits go from 0 to \( R \), capturing the entire charge distribution.
• Spherical coordinates are used, as they're ideal for spherical symmetry.

This method allows us to calculate the total charge contained within the sphere, ensuring all regions where the density is non-zero are taken into account.

Gauss's Law

Gauss's Law is a fundamental principle that relates the electric field flux through a closed surface to the charge enclosed by that surface. In mathematical terms, Gauss's Law is written as:\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{encl}}}{\varepsilon_0} \]Gauss's Law is particularly effective for problems with high symmetry, such as spherical symmetry, making it easier to calculate electric fields generated by symmetric charge distributions.
In this exercise, Gauss's Law is applied to both regions:

• For \( r \geq R \), the electric field can be treated as if all the charge were located at the center of the sphere. This simplifies calculations considerably.
• For \( r \leq R \), the electric field is calculated from the charge inside a Gaussian surface of radius \( r \), ensuring only the charge enclosed by this surface contributes to the electric field at \( r \).

Gauss's Law enables us to efficiently find expressions for the electric field in different regions of this problem.

Spherical Symmetry

Spherical symmetry is a condition where a system looks the same from every direction around a central point. In the context of this problem, spherical symmetry simplifies the calculation of electric fields and charge distribution.
The key aspect of spherical symmetry is that properties only depend on the radial distance \( r \) from the center, not the specific direction:

• It means the electric field vector at a point only has a radial component, simplifying how we apply Gauss's Law.
• In spherical coordinates, the choice of surface (like a sphere) to evaluate Gauss's Law becomes natural because of the symmetry of the problem.

Using spherical symmetry, we see how to effectively apply mathematical tools to derive the electric field, emphasizing this concept's importance in reducing computational complexity.

Maximum Electric Field

Finding the maximum value of the electric field involves determining the point within the region \( r \leq R \) where the electric field intensity is greatest.
This involves calculus:

• The electric field inside the sphere is expressed as \( E(r) = \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{r^2}{R} \right) \).
• Taking the derivative \( \frac{dE}{dr} \) helps find where the electric field has a maximum. This point occurs at \( r = \frac{R}{2} \).
• Substituting \( r = \frac{R}{2} \) back into \( E(r) \) gives us the maximum electric field value \( E_{\text{max}} = \frac{\rho_0 R}{12 \varepsilon_0} \).

Finding maxima in physics involves using derivatives to pinpoint key points of interest, showing how calculus can be invaluable in understanding fields and forces.