Problem 66 - University Physics with Modern Physics

Question

\(22.66 .\) 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 $$ \begin{array}{ll}{\rho(r)=\alpha} & {\text { for } r \leq R / 2} \\\ {\rho(r)=2 \alpha(1-r / R)} & {\text { for } R / 2 \leq r \leq R} \\\ {\rho(r)=0} & {\text { for } r \geq R}\end{array} $$ 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 \(\vec{E}\) as a function of \(r .\) Do this separately for all three regions. Express your answers in terms of the total charge \(Q\) . Be sure to check that your results agree on the boundaries of the regions. (c) What fraction of the total charge is contained within the region \(r \leq R / 2 ?\left(\text { d) If an electron with charge } q^{\prime}=-e \text { is }\right.\) oscillating back and forth about \(r=0\) (the center of the distribution) with an amplitude less than \(R / 2,\) show that the motion is simple harmonic. (Hint: Review the discussion of simple harmonic motion in Section 13.2. If, and only if, the net force on the electron is proportional to its displacement from equilibrium, then the motion is simple harmonic. \()(\mathrm{e})\) What is the period of the motion in part \((\mathrm{d}) ?(\mathrm{f})\) If the amplitude of the motion described in part (e) is not?

Step-by-Step Solution

Answer

\( \alpha = \frac{3Q}{\pi R^3} \) and \( E(r) \) is found using Gauss's Law. Motion is simple harmonic with period \( T = 2\pi \sqrt{\frac{3m\varepsilon_0}{e\alpha}} \).

Step 1: Understanding Charge Distribution

The volume charge density \( \rho(r) \) is piecewise defined: constant \( \alpha \) for \( r \leq R/2 \), decreases linearly for \( R/2 \leq r \leq R \), and is zero for \( r \geq R \). The positive charge \( Q \) is spread over radius \( R \). The aim is to find \( \alpha \) in terms of \( Q \) and \( R \).

Step 2: Solve for \( \alpha \) in terms of \( Q \) and \( R \)

Calculate \( Q \) by integrating \( \rho(r) \) over the spherical volume:\[Q = \int_0^R \rho(r) \, 4\pi r^2 \, dr \]Calculate separately for the intervals: \( 0 \leq r \leq R/2 \) and \( R/2 \leq r \leq R \).For \( r \leq R/2 \):\[Q_1 = \int_0^{R/2} \alpha \, 4\pi r^2 \, dr = \frac{4\pi\alpha}{3} \left(\frac{R}{2}\right)^3 \]For \( R/2 \leq r \leq R \):\[Q_2 = \int_{R/2}^{R} 2\alpha(1-r/R) \, 4\pi r^2 \, dr \]You solve this integral to find \( Q_2 \).Then set \( Q = Q_1 + Q_2 \) and solve for \( \alpha \).

Step 3: Apply Gauss's Law to Find \( \vec{E}(r) \)

Use Gauss's Law \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{ ext{enc}}}{\varepsilon_0} \) to find \( \vec{E}(r) \) for each region. For \( r \leq R/2 \), \( Q_{\text{enc}} = \frac{4\pi\alpha}{3}r^3 \), so:\[E(r) = \frac{\alpha r}{3\varepsilon_0}\]For \( R/2 \leq r \leq R \), calculate \( Q_{\text{enc}} \) using the previous integrals and apply Gauss's Law:\[E(r) = \frac{1}{4\pi\varepsilon_0} \cdot \left(\text{Expression in terms of } Q, R, \text{ and } r \right)\]For \( r \geq R \), all charge is enclosed:\[E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}\]

Step 4: Charge Fraction for \( r \leq R/2 \)

The charge fraction is \( Q_1/Q \). Since \( Q_1 \) was found in Step 2, set \( \text{Charge Fraction} = \frac{Q_1}{Q} = \text{Rational in terms of } \frac{R}{2} \).

Step 5: Analyze Simple Harmonic Motion for Electron

Show the force on an electron within \( r \leq R/2 \) is proportional to displacement:\[F = q'E = -e\cdot \frac{\alpha r}{3\varepsilon_0}\]Since force \( F \propto -r \), it is simple harmonic.

Step 6: Finding the Period of Oscillation

From \( F = -kx \), where \( k = \frac{e\alpha}{3\varepsilon_0} \), use the formula for period \( T = 2\pi \sqrt{\frac{m}{k}} \). Determine \( T \).Then substitute all constants and express \( T \) explicitly.

Key Concepts

Spherical Charge Distribution

In the exercise, we encounter a spherical charge distribution described by a unique piecewise function. This means the charge density changes based on the region we are in within the sphere. Here's a breakdown of the regions:

• For radius less than or equal to \(R/2\), the charge density is constant at \(\alpha\).
• For the region \(R/2 \leq r \leq R\), the density decreases linearly. This linear decrease is given by the expression \(2\alpha(1 - r/R)\).
• At radii greater than \(R\), the charge density is zero, meaning there’s no charge present.

This distribution is important because it influences how we calculate quantities like the total charge \(Q\). It requires integrating the expressions for different regions, taking into account their unique functions. Understanding this can help when calculating fields influenced by these charges.

Simple Harmonic Motion

Simple harmonic motion (SHM) is a type of oscillatory motion. A key component is that the force acting on the object is proportional to the displacement from its equilibrium position and acts in the opposite direction. In this exercise, we consider an electron oscillating within a charge distribution.
Here's how SHM is applied:

• The electron, with charge \(-e\), experiences a force when placed inside the charged sphere, where the charge density is constant (for \(r \leq R/2\)).
• The electric force is calculated through the formula \(F = q'E = -e \cdot \frac{\alpha r}{3\varepsilon_0}\), showing us the force is proportional to \(-r\).

Since the force is proportional to the displacement \(r\), it confirms the motion is simple harmonic. This results in oscillations back and forth about the center, like a spring-mass system. This example mirrors the basics of SHM typically discussed within physics, where restoring forces create stable periodic movement.

Electric Field Calculation

Electric field calculations within this context often revolve around Gauss's Law. Gauss's Law states that the electric field around a closed surface is related to the charge enclosed by that surface. Using this law simplifies finding electric fields in symmetric charge distributions.
Here's how it works for our spherical distribution:

• In the region \(r \leq R/2\), the field, \(E(r)\), depends on the enclosed charge. By using \(Q_{\text{enc}} = \frac{4\pi\alpha}{3}r^3\), Gauss's Law gives \(E(r) = \frac{\alpha r}{3\varepsilon_0}\).
• For \(R/2 \leq r \leq R\), additional calculations modify \(Q_{\text{enc}}\), incorporating the linear decrease in charge density, which then affects \(E(r)\).
• Beyond \(R\), all the charge is enclosed, leading to the familiar form \(E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}\) of the electric field similar to that outside a sphere with charge \(Q\).

These calculations show how crucial it is to understand both the distribution of charge and how to apply Gauss's Law effectively. The law allows us to bypass complex integrations by providing solutions that depend solely on symmetry and enclosed charge.