Problem 53 - University Physics with Modern Physics

Question

A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho\)(\(r\)) given as follows: $$\rho(r) = \rho_0 \bigg(1 - \frac{r}{R}\bigg) \space \space \space \mathrm{for} \space r \leq R$$ $$\rho(r) = 0 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \mathrm{for} \space r \leq R$$ where \(\rho_0 = 3Q/{\pi}R^3\) is a positive constant. (a) Show that the total charge contained in the charge distribution is \(Q\). (b) Show that the electric field in the region \(r \geq R\) is identical to that produced by a point charge \(Q\) at \(r =\) 0. (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.

Step-by-Step Solution

Answer

Total charge is \( Q \). The external electric field is \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \). Inside, the maximum electric field occurs at \( r = \frac{R}{3} \) with value \( E_{max} = \frac{Q}{6\pi \varepsilon_0 R^2} \).

Step 1: Understanding Charge Density

The given charge density for a spherically symmetric charge distribution is \( \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \) for \( r \leq R \), and \( \rho(r) = 0 \) for \( r > R \). Observe that \( \rho_0 = \frac{3Q}{\pi R^3} \). This distribution follows a radial decay from the center.

Step 2: Calculate Total Charge

To find the total charge \( Q \), integrate the charge density over the volume of the sphere: \( Q = \int_0^R \rho(r) \cdot 4\pi r^2 \cdot dr \). Substitute: \( \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \) and calculate the volume integral.

Step 3: Integration for Total Charge

Integrate \( Q = \int_0^R \rho_0 \left(1 - \frac{r}{R}\right) 4\pi r^2 \, dr \). Simplify the integral to: \( Q = 4\pi \rho_0 \left(\int_0^R r^2 \ dr - \frac{1}{R} \int_0^R r^3 \, dr \right) \). Evaluate to find \( Q = \frac{4\pi \rho_0 R^3}{3} \). Substitute \( \rho_0 = \frac{3Q}{\pi R^3} \) to show that \( Q = Q \).

Step 4: Show Electric Field Outside

According to Gauss's Law, for \( r \geq R \), the electric field behaves as if all charge was concentrated at a point at the center. The field is \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \). This is identical to any other point charge \( Q \) centered at \( r = 0 \) for \( r \geq R \).

Step 5: Calculate Electric Field Inside

Inside the sphere \( (r \leq R) \), use Gauss's Law: \( E \cdot 4\pi r^2 = \frac{1}{\varepsilon_0} \cdot \int_0^r \rho(r') \cdot 4\pi {r'}^2 \, dr' \). The integral becomes: \( E \cdot 4\pi r^2 = \frac{3Q}{R^3 \varepsilon_0} \left( \frac{r^3}{3} - \frac{r^4}{4R} \right) \). Solve for \( E \).

Step 6: Expression for Electric Field Inside

After simplification, the electric field inside is given by: \( E = \frac{Qr}{4\pi \varepsilon_0 R^3} \left(2 - \frac{3r}{R}\right) \). This expression is valid for \( r \leq R \).

Step 7: Plot Electric Field

Graph \( E \) as a function of \( r \). For \( r \leq R \), use the expression: \( E = \frac{Qr}{4\pi \varepsilon_0 R^3} \left(2 - \frac{3r}{R}\right) \). For \( r > R \), use \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \). The graph should show a peak inside the sphere and decrease with \( r^{-2} \) outside.

Step 8: Finding Maximum Electric Field

To find the maximum field inside \( r \leq R \), differentiate \( E(r) = \frac{Qr}{4\pi \varepsilon_0 R^3} \left(2 - \frac{3r}{R}\right) \) with respect to \( r \) and set the derivative to zero. This gives the maximum field condition: \( 2 - \frac{6r}{R} = 0 \), solving gives \( r = \frac{R}{3} \).

Step 9: Maximum Electric Field Value

Substitute \( r = \frac{R}{3} \) back into \( E(r) \): \( E_{max} = \frac{Q}{4\pi \varepsilon_0 R^3} \left(\frac{R}{3}\right) \left(2 - 1 \right) = \frac{Q}{6\pi \varepsilon_0 R^2} \).

Key Concepts

Charge Distribution

Charge distribution refers to how electric charges are spread out in space. In this exercise, we are dealing with a non-uniform charge distribution that is also spherically symmetric. This means that the charge density, or the amount of charge per unit volume, varies with the distance from the center of the sphere but is the same in every direction from that center.

The charge density \( \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \) decreases linearly from the center, reaching zero at the surface of the sphere. Beyond the radius \( R \), the charge density becomes zero. This implies that all the charge is concentrated within the sphere of radius \( R \).

Understanding charge distribution is crucial because it determines how the electric field behaves both inside and outside the sphere. A non-uniform charge distribution will affect the field differently compared to a uniform one, leading to variations in electric field strength at different points.

Gauss's Law

Gauss's Law is a fundamental principle that relates the electric field to the distribution of electric charge. The law states that the electric flux through a closed surface is proportional to the charge enclosed by the surface. Mathematically, it is given by:

\[ \Phi = \oint E \cdot dA = \frac{Q_{enc}}{\varepsilon_0} \]

Where \( \Phi \) is the electric flux, \( E \) is the electric field, \( dA \) is a differential area on the closed surface, \( Q_{enc} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space.

In the context of the problem, Gauss's Law helps us determine the electric field both inside and outside the sphere. For \( r \geq R \), we apply Gauss’s Law to a spherical Gaussian surface that encompasses all the charge \( Q \). The field behaves as if all charge is concentrated at the sphere's center. Inside the sphere, the law is used to account for the charge enclosed within a smaller sphere of radius \( r \). This analysis enables us to derive the expression for the electric field at any point within the sphere.

Spherical Symmetry

Spherical symmetry is a situation where a system looks the same when viewed from any direction around its center. In this problem, the spherical symmetry of the charge distribution simplifies the analysis of the electric field. Because of this symmetry, the electric field at any point outside a spherically symmetric charge distribution depends only on the radial distance from the center and not on the direction.

When using spherical symmetry and Gauss's Law together, you can assume a symmetrical Gaussian surface like a sphere to simplify calculations. This is possible because of the symmetry of the system, leading to a constant electric field magnitude on the surface, making the integration straightforward. The resulting electric field expression applies uniformly in any direction from the center, significantly simplifying how we approach finding the electric field in regions inside and outside the sphere.

Charge Density

Charge density is a measure of how much electric charge is present in a particular space. It is usually expressed in terms of charge per unit volume, \( \rho(r) = \frac{dq}{dV} \). Understanding charge density is crucial because it directly influences the electric force at any point within or around the charged object.

In our exercise, the charge density \( \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \) is a function of the radial distance \( r \). This function decreases linearly from the center of the distribution, reflecting a non-uniform distribution. Such a distribution causes a change in the internal electric field variation compared to a uniform one.

Charge density calculations often involve integrating this function over a volume to find the total charge. For example, within the given sphere, integrating the charge density helps us verify that the total charge sums to \( Q \). Comprehending how charge density modifies across the system reveals insights into the resulting electric field dynamics, aiding in efficient analysis of the problem at hand.