Problem 62 - University Physics with Modern Physics

Question

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 $$\rho(r) = 3ar/2R \space \space \space \space \space \mathrm{for} \space r \leq R/2$$ $$\rho(r) = \alpha[1-(r/R)^2] \space \space \space \space \mathrm{for} \space R/2 \leq r \leq R$$ $$\rho(r) = 0 \space \space \space \space \space \space \mathrm{for} \space r \geq R$$ Here \(\alpha\) is a positive constant having units of C/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 \(Q\). (c) What fraction of the total charge is contained within the region \(R/2 \leq r \leq R\)? (d) What is the magnitude of \(\overrightarrow{E}\) at \(r = R/2\)? (e) If an electron with charge \(q' = -e\) is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why?

Step-by-Step Solution

Answer

Determine charge distribution, solve electric field in each region, and evaluate specific conditions.

Step 1: Determine expression for total charge

First, we need to find an expression for the total charge contained within the sphere using the given charge density expressions. The charge density has different expressions in two regions, so we need to integrate separately for each region.For the inner region where \( r \leq R/2 \), the charge \( dQ_1 \) is given by:\[ dQ_1 = \rho(r) \, dV = \frac{3ar}{2R} \cdot 4\pi r^2 \, dr \]Integrate from 0 to \( R/2 \).For the outer region where \( R/2 \leq r \leq R \), the charge \( dQ_2 \) is given by:\[ dQ_2 = \alpha \left[1 - \left(\frac{r}{R}\right)^2\right] 4\pi r^2 \, dr \]Integrate from \( R/2 \) to \( R \).The total charge is the sum:\[ Q = Q_1 + Q_2 \]

Step 2: Calculate \( Q_1 \) and \( Q_2 \)

**Calculation of \( Q_1 \):**\[Q_1 = \int_0^{R/2} \frac{3ar}{2R} \cdot 4\pi r^2 \, dr = \frac{6\pi a}{2R} \int_0^{R/2} r^3 \, dr\]This evaluates to:\[Q_1 = \frac{3\pi a}{R} \left(\frac{r^4}{4}\right)\bigg|_0^{R/2} = \frac{3\pi a}{R} \cdot \frac{R^4}{256} = \frac{3\pi a R^3}{128}\]**Calculation of \( Q_2 \):**\[Q_2 = \alpha \cdot 4\pi \int_{R/2}^{R} r^2 \left[1 - \left(\frac{r}{R}\right)^2\right] \, dr\]Use the substitution \( x = r/R \), thus \( dx = dr/R \).\[Q_2 = \alpha \cdot 4\pi R^3 \int_{1/2}^{1} x^2 \left(1 - x^2\right) \, dx = \alpha \cdot 4\pi R^3 \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_{1/2}^1\]Evaluating this gives:\[Q_2 = \alpha \cdot 4\pi R^3 \left(\frac{1}{3} - \frac{1}{5} - \left(\frac{1}{24} - \frac{1}{80}\right)\right)\]

Step 3: Solve for \( \alpha \) in terms of \( Q \) and \( R \)

From Step 2, the expressions for \( Q_1 \) and \( Q_2 \) allow us to express the total charge as:\[Q = \frac{3\pi a R^3}{128} + \alpha \cdot 4\pi R^3 \left(\frac{2}{15} - \frac{1}{160}\right)\]To solve for \( \alpha \):- Set the equation equal to \( Q \) and solve for \( \alpha \).- \(\alpha = \frac{Q - \frac{3\pi a R^3}{128}}{4\pi R^3 \left(\frac{31}{240}\right)}\) Simplify to get an expression of \( \alpha \).

Step 4: Use Gauss's law to derive electric field (Region 1)

In the innermost region \( r \leq R/2 \), use Gauss's law:\[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}\]Since \( E \) is uniform on a Gaussian sphere of radius \( r \leq R/2 \),\[E \, 4\pi r^2 = \frac{1}{\varepsilon_0} \int_0^r \frac{3ar}{2R} \, 4\pi r^2 \, dr\]Calculating the integral,\[E(r) = \frac{1}{\varepsilon_0} \cdot \frac{3ar^4}{8R}\]

Step 5: Use Gauss's law to derive electric field (Region 2)

For the region \( R/2 \leq r \leq R \), the enclosed charge is the sum of the charge in the inner region and part of the outer region.Use Gauss's law:\[E \, 4\pi r^2 = \frac{1}{\varepsilon_0} \left(\frac{3\pi a R^3}{128} + \int_{R/2}^r \alpha[1-(x/R)^2] 4\pi x^2 \, dx\right)\]Solving this:- Derive expression using integral.- Calculate \( E(r) \) which will involve keeping track of both terms.

Step 6: Use Gauss's law to derive electric field (Region 3)

For the outermost region where \( r \geq R \), the entire charge is enclosed, so:\[E \, 4\pi r^2 = \frac{Q}{\varepsilon_0}\]Thus the electric field is given by:\[E(r) = \frac{Q}{4\pi \varepsilon_0 r^2}\]

Step 7: Calculate fraction of total charge in \( R/2 \leq r \leq R \)

The charge in the region \( R/2 \leq r \leq R \) is given by \( Q_2 \), which we calculated earlier.The fraction is:\[\frac{Q_2}{Q}\]Substitute the expression of \( Q_2 \) and solve.

Step 8: Determine electric field at \( r = R/2 \)

Substitute \( r = R/2 \) into the expression of \( E(r) \) derived for \( r \leq R/2 \):\[E\left(\frac{R}{2}\right) = \frac{1}{\varepsilon_0} \cdot \frac{3a (R/2)^4}{8R}\]This simplifies to give the magnitude of the electric field at \( r = R/2 \).

Step 9: Explain the motion of an electron in the regions

Even though the motion is oscillatory, it is not simple harmonic because of the non-linear electric field dependency on distance \( r \). Simple harmonic motion requires that the restoring force is directly proportional to \( -r \), which is not the case in any of the regions given the derived field expressions.

Key Concepts

Electric Field

The concept of the electric field is essential when dealing with charged objects. An electric field represents the region around a charged body where its force can be felt by other charges. It is a vector field, typically denoted by \(\vec{E}\), and is defined as the force per unit charge exerted on a small positive test charge placed in the field.

According to Gauss's Law, which is a fundamental principle in electromagnetism, the electric field can be determined by calculating the net charge enclosed within a given surface. For spherically symmetric charge distributions, as in our exercise, this simplifies the evaluation of electric fields considerably. Using a spherical Gaussian surface, the problem becomes solvable by evaluating the symmetry.

• In the innermost region \( r \leq R/2 \), the field is calculated by considering the charge within the sphere of radius \( r \).
• In the region \( R/2 \leq r \leq R \), the field depends both on the inner core charge and on the additional charge in the shell.
• Finally, outside the entire distribution \( r \geq R \), the total charge \( Q \) is enclosed. The electric field behaves like that of a point charge.

This step-by-step application of Gauss's Law simplifies finding the electric field across different regions of a spherically charged distribution.

Volume Charge Density

Volume charge density, denoted as \( \rho(r) \), describes how charge is distributed within a given volume. It represents the amount of charge per unit volume at any point within that space. In this exercise, \( \rho(r) \) is given by different expressions depending on the region within the sphere:

• For \( 0 \leq r \leq R/2 \), \( \rho(r) = \frac{3ar}{2R} \) indicates an increasing charge density with radius.
• For \( R/2 \leq r \leq R \), \( \rho(r) = \alpha[1-(r/R)^2] \) shows a density decreasing as we approach the outer surface.
• Beyond \( R \), the density is zero, meaning no charge exists outside the sphere.

Understanding these variations is crucial, as they determine the electric field in each region. This non-uniform distribution requires careful integration of the density across the volume to determine the total charge.

Spherical Charge Distribution

A spherical charge distribution assigns how charges are arranged within a sphere. This type of distribution is particularly interesting because of its symmetry, which simplifies the application of Gauss's Law.

In this situation, by dividing the sphere into different regions, each with its specific volume charge density, we can better analyze how the charge contributes to the electric field. In general, spherical charge distributions can have:

• Homogeneous distribution where the charge density is constant throughout the volume.
• Varying distribution, as seen here, with different expressions of charge density depending on the region.

By using integration, we compute the charge contained in varying sections of the sphere, which is fundamental for later finding the constants like \( \alpha \) in terms of total charge \( Q \) and the sphere's radius \( R \).
This understanding also aids in determining how charges are configured and how they influence phenomena such as electric fields and potential within and outside the sphere.