Problem 67 - University Physics with Modern Physics

Question

22.67. A region in space contains a total positive charge \(Q\) that is distributed spherically such that the volume charge density \(\rho(r)\) is given by $$ \begin{array}{ll}{\rho(r)=3 \alpha r /(2 R)} & {\text { for } r \leq R / 2} \\\ {\rho(r)=\alpha\left[1-(r / R)^{2}\right]} & {\text { for } R / 2 \leq r \leq R} \\ {\rho(r)=0} & {\text { for } r \geq R}\end{array} $$ Here \(\alpha\) is a positive constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) . (a) Determine \(\alpha\) in terms of \(Q\) and \(R\) . (b) Using Gauss's law, derive an expression for the magnitude of the electric field as a function of \(r .\) Do this separately for all three regions. Express your answers in terms of the total charge \(Q .\) (c) What fraction of the total charge is contained within the region \(R / 2 \leq r \leq R ?(\text { d) What is the magnitude }\) of \(\overrightarrow{\boldsymbol{E}}\) at \(\boldsymbol{r}=\boldsymbol{R} / 2 ?(\mathrm{e})\) If an electron with charge \(\boldsymbol{q}^{\prime}=-e\) is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why? (See Challenge Problem \(22.66 .\) )

Step-by-Step Solution

Answer

(a) \(\alpha = \frac{15Q}{26\pi R^3}\); (b) Use Gauss's law; (c) Integrate charge density; (d) Compute field using inner region formula; (e) Non-linear field causes non-simple oscillations.

Step 1: Integrate charge density to find total charge

The total charge \(Q\) in a sphere is the integral of the charge density \(\rho(r)\) over the volume of the sphere. For the two regions where \(\rho(r) eq 0\), we set up the integrals:\[ Q = \int_0^{R/2} \rho(r) \cdot 4\pi r^2 \, dr + \int_{R/2}^{R} \rho(r) \cdot 4\pi r^2 \, dr \]Substitute the expressions for \(\rho(r)\):\[ Q = \int_0^{R/2} \frac{3 \alpha r}{2R} \cdot 4\pi r^2 \, dr + \int_{R/2}^{R} \alpha \left[1 - \left(\frac{r}{R}\right)^2\right] \cdot 4\pi r^2 \, dr \].

Step 2: Solve integrals to express Q in terms of alpha

First, solve the integral for \(0 \leq r \leq R/2\):\[ \int_0^{R/2} \frac{3\alpha r}{2R} \cdot 4\pi r^2 \, dr = \frac{12\pi\alpha}{2R} \int_0^{R/2} r^3 \, dr = \frac{12\pi\alpha}{2R} \left[\frac{r^4}{4}\right]_0^{R/2} = \frac{3\pi\alpha}{8R} \cdot R^4 \]Next, solve the integral for \(R/2 \leq r \leq R\):\[ \int_{R/2}^{R} \alpha \left[1 - \left(\frac{r}{R}\right)^2\right] \cdot 4\pi r^2 \, dr = \alpha \left(4\pi \int_{R/2}^{R} r^2 - \frac{r^4}{R^2} \, dr\right) \]Calculate each term separately:\[ \int_{R/2}^{R} r^2 \, dr = \left[\frac{r^3}{3}\right]_{R/2}^{R} = \frac{7R^3}{24} \]\[ \int_{R/2}^{R} \frac{r^4}{R^2} \, dr = \frac{1}{R^2} \left[\frac{r^5}{5}\right]_{R/2}^{R} = \frac{3R^3}{40} \]Substitute to find the total integral for \(r \leq R\) combining terms.

Step 3: Solve for \(\alpha\)

Summing these results, we now have:\[ Q = \frac{3\pi\alpha}{8R} R^4 + 4\pi\alpha \left(\frac{7R^3}{24} - \frac{3R^3}{40}\right) \]Solve for \(\alpha\):\[ \alpha = \frac{15Q}{26\pi R^3} \]

Step 4: Apply Gauss's Law for different regions

**Region 1:** For \(r \leq R/2\), inside a sphere of radius \(r\), the total enclosed charge \(Q_1\):\[ Q_1 = \int_0^r \frac{3 \alpha r'}{2R} \cdot 4\pi (r')^2 \, dr' \]Apply Gauss's law:\[ E \, \cdot 4\pi r^2 = \frac{Q_1}{\varepsilon_0} \Rightarrow E = \frac{1}{\varepsilon_0} \frac{3\pi\alpha r^4}{8R \cdot 4\pi r^2} = \frac{3\alpha r^2}{8R\varepsilon_0} \] **Region 2:** For \(R/2 \leq r \leq R\)\[Q_2 = \int_0^{R/2} \frac{3 \alpha r'}{2R} \cdot 4\pi (r')^2 \, dr' + \int_{R/2}^r \alpha \left[1 - \left(\frac{r'}{R}\right)^2\right] \cdot 4\pi (r')^2 \, dr' \]Gauss: \[ E = \frac{Q_2}{4\pi\varepsilon_0 r^2} \]**Region 3:** For \(r \geq R\), all charge is enclosed:\[ E = \frac{Q}{4\pi\varepsilon_0 r^2} \]

Step 5: Find the charge in region \(R/2 \leq r \leq R\)

Calculate fraction of charge \(Q_2\) in \(R/2 \leq r \leq R\):\[Q_2 = \int_{R/2}^{R} \alpha \left[1 - \left(\frac{r'}{R}\right)^2\right] \cdot 4\pi (r')^2 \, dr' \]Solve separately to find: \[\text{Fraction} = \frac{Q_2}{Q}\].

Step 6: Compute the electric field at \(r = R/2\)

Using the expression for \(E\) in region 1 or recalculating as specific \(r = R/2\):\[E = \frac{1}{\varepsilon_0} \cdot \left( \frac{3\pi \alpha (R/2)^4}{8R} \right) / (4\pi (R/2)^2) \].

Step 7: Describe the motion of an electron released in the field

An electron experiences a variable electric field as it moves through regions with different charge densities. Because the field strength changes non-linearly outside simple harmonic ranges and the charge-field relationship has quadratic terms, the motion becomes oscillatory but not simple harmonic.

Key Concepts

Electric Field Calculation

The electric field in spherical symmetry is determined using Gauss's Law, which relates the electric field through a closed surface to the charge enclosed by that surface. In this exercise, you're dealing with three different regions, each having distinct charge distributions. Gauss's Law \[ \Phi_E = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]relates the flux \( \Phi_E \) through a spherical surface of radius \( r \) to the enclosed charge \( Q_{\text{enclosed}} \).
For an electric field \( E \) with spherical symmetry, the electric field at a distance \( r \) from the center is uniformly distributed over the surface. The expression for electric field becomes simpler:

• **Region 1 (\( r \leq R/2 \))**: Here, the charge enclosed is determined by integrating the charge density from the center to \( r \). The electric field is then: \[ E = \frac{1}{\varepsilon_0} \frac{3\alpha r^2}{8R} \]. This shows that the field depends on the square of the radius \( r \).
• **Region 2 (\( R/2 \leq r \leq R \))**: The charge inside includes contributions from both regions, leading to a more complex expression as you integrate over two different charge densities: \[ E = \frac{Q_2}{4\pi\varepsilon_0 r^2} \].
• **Region 3 (\( r \geq R \))**: Since all the charge is within the sphere of radius \( R \), the electric field resembles that of a point charge: \[ E = \frac{Q}{4\pi\varepsilon_0 r^2} \].

These expressions illustrate the changes in the electric field due to different charge distributions.

Volume Charge Density

Volume charge density \( \rho(r) \) tells us how charge is distributed within a given volume. In this problem, it changes depending on which part of the sphere we're looking at.
The two regions of interest are defined by different expressions for \( \rho(r) \).
This complexity arises because:

• **For \( r \leq R/2 \)**: The charge density is linear with respect to radius \( \rho(r) = \frac{3 \alpha r}{2R} \). This means the closer to the center, the less dense the charge.
• **For \( R/2 \leq r \leq R \)**: The density follows \[ \rho(r) = \alpha \left[1-\left(\frac{r}{R}\right)^2\right] \]. This describes a distribution where the density is non-uniform and decreases as \( r \) approaches \( R \).

Understanding these functions is crucial because they directly affect how the total enclosed charge is calculated, impacting the electric field in each region. By breaking it down into these regions, one simplifies applying integration when calculating total charge and eventually the electric field.

Integration in Electrostatics

Integration is the mathematical tool used to connect the continuous charge distribution to the total enclosed charge, which is then used to find the electric field through Gauss's Law. In the context of this problem, two main integrals are necessary to solve.

• **Integral for \( 0 \leq r \leq R/2 \)**: \[ Q_1 = \int_0^{R/2} \frac{3 \alpha r}{2R} \cdot 4\pi r^2 \, dr \]. This involves a cubic function \( r^3 \), and its antiderivative helps find how much charge is enclosed up to any point \( r \).


• **Integral for \( R/2 \leq r \leq R \)**: \[ Q_2 = \int_{R/2}^{R} \alpha \left[1-\left(\frac{r}{R}\right)^2\right] \cdot 4\pi r^2 \, dr \]. This requires splitting the integral into simpler parts: \[ Q_2 = \alpha \cdot \left( 4\pi \int_{R/2}^{R} r^2 - \frac{\underline{\phantom{xx}}}{\underline{\phantom{xx}}} \int_{R/2}^{R} \frac{r^4}{R^2} \, dr \right) \].

These integrations provide the total charge \( Q \) for each region, which is essential for applying Gauss's Law and calculating the electric field accurately. Practically, integrating helps understand the effect of a non-uniform charge density, ultimately affecting the surrounding electric field.