Problem 32 - Fundamentals of Physics

Question

(so \(\mathrm{A}\) long, nonconducting, solid cylinder of radius \(4.0 \mathrm{~cm}\) has a nonuniform volume charge density \(\rho\) that is a function of radial distance \(r\) from the cylinder axis: \(\rho=\) \(A r^{2}\). For \(A=2.5 \mu \mathrm{C} / \mathrm{m}^{5}\), what is the magnitude of the electric field at (a) \(r=3.0 \mathrm{~cm}\) and (b) \(r=5.0 \mathrm{~cm} ?\)

Step-by-Step Solution

Answer

At \( r = 3.0 \text{ cm} \), calculate the electric field using \( E = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 rh} \) with enclosed charge integral. At \( r = 5.0 \text{ cm} \), calculate using the total charge of the cylinder with \( E = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 \cdot 0.05 \cdot h} \).

Step 1: Understand the Problem

We have a nonconducting solid cylinder with a volume charge density given by \( \rho = A r^2 \). We need to find the electric field at specific radial distances \( r \), both inside and outside the cylinder. The radius of the cylinder is \( R = 4.0 \text{ cm} \).

Step 2: Express Gauss's Law for a Cylinder

Gauss's law relates the electric field to the charge enclosed within a Gaussian surface. For a cylindrical surface, it can be written as \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_{0}} \), where \( \varepsilon_{0} \) is the permittivity of free space.

Step 3: Calculate Charge Enclosed for \(r < R\)

For \( r = 3.0 \text{ cm} \), which is inside the cylinder, the charge enclosed \( Q_{\text{enc}} \) is the integral of \( \rho \) over a volume of height \( h \): \[ Q_{\text{enc}} = \int_0^r \rho \cdot 2\pi rh \, dr = \int_0^{0.03} 2.5 \times 10^{-6} r^2 \cdot 2\pi rh \, dr \]Solve the integral to find \( Q_{\text{enc}} \).

Step 4: Solve for Electric Field at \(r = 3.0 \text{ cm}\)

Use Gauss's Law to find \( E \) inside the cylinder: \[ 2\pi rh E = \frac{Q_{\text{enc}}}{\varepsilon_{0}} \]Simplify to get: \[ E = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 rh} \]

Step 5: Calculate Charge Enclosed for \(r > R\)

For \( r = 5.0 \text{ cm} \), which is beyond the cylinder, the charge enclosed \( Q_{\text{enc}} \) is the total charge of the cylinder (only out to \( R \)). Evaluate:\[ Q_{\text{enc}} = \int_0^R \rho \cdot 2\pi rh \, dr = \int_0^{0.04} 2.5 \times 10^{-6} r^2 \cdot 2\pi rh \, dr \]Solve this integral.

Step 6: Solve for Electric Field at \(r = 5.0 \text{ cm}\)

Apply Gauss's Law considering \( r > R \):\[ 2\pi rh E = \frac{Q_{\text{enc}}}{\varepsilon_{0}} \]Solve:\[ E = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 \cdot 0.05 \cdot h} \]

Step 7: Solve the Integrals and Simplify

Evaluate the integrals from Step 3 and Step 5. For \( r = 3.0 \text{ cm} \), \[ Q_{\text{enc}} = \int_0^{0.03} 2.5 \times 10^{-6} r^2 \cdot 2\pi rh \, dr \]For \( r = 5.0 \text{ cm} \),\[ Q_{\text{enc}} = \int_0^{0.04} 2.5 \times 10^{-6} r^2 \cdot 2\pi rh \, dr \]Substitute back into the equations for \( E \).

Key Concepts

Electric field in nonconducting materials

In nonconducting materials, such as the solid cylinder in our problem, electric fields behave differently than in conductors. The electric field originates from the volume charge density within the material. Unlike conductors, where charges reside on the surface, nonconductors or insulators allow charge distribution throughout their volume.

The field within and outside a nonconducting object is determined by Gauss's Law. According to this law, the electric flux through a closed surface is proportional to the charge enclosed. This means that if we want to find the electric field produced by a uniformly charged nonconductor, such as our cylinder, we examine the symmetry of the charge distribution and derive the field using suitable Gaussian surfaces. This approach works for various shapes, but here we consider the special case of a cylinder.

It's crucial to note that in nonconductors, the internal electric field isn't zero as it might be in conductors. The field varies based on how charge is distributed inside the material, influencing both the magnitude and direction of the field at different points.

Volume charge density

Volume charge density is a measure of how charge is distributed in a given volume, denoted by the symbol \( \rho \). For our cylindrical problem, the charge density is not constant; instead, it varies with the radial distance from the axis, given by \( \rho = A r^2 \).

This function indicates that the density increases with the square of the distance from the center. This variance means that more charge is packed into outer regions than near the axis. This nonuniform distribution is what leads us to calculate the electric field differently from that in uniform charge distributions. The specific form of \( \rho \) dictates how we integrate over the volume to find the total enclosed charge when applying Gauss's Law.

The concept of volume charge density is critical because it establishes the origin of the electric field within a material. Depending on the setup, this density can be constant or a function of position, significantly affecting our calculations.

Cylindrical symmetry

Cylindrical symmetry plays a pivotal role in simplifying the electric field calculations around our problem's cylindrical structure. When a system exhibits cylindrical symmetry, all physical quantities depend only on the radial distance from the axis and not on the angular position or the height along the cylinder.

This symmetry simplifies the application of Gauss's Law. We choose a cylindrical Gaussian surface that aligns with the symmetrical properties of the charge distribution. As a result, we don't need to consider variations along the angular or height dimensions of the cylinder. This alignment allows us to focus strictly on the radial components to find the electric field.

By exploiting symmetry, we can convert complex integral calculus into simpler arithmetic, isolating the variables we need to obtain straightforward expressions for the electric field at different locations. Cylindrical symmetry thus becomes a crucial tool in reducing the complexity of a problem.

Permittivity of free space

The permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental physical constant that characterizes how electric fields interact with the vacuum. Its value is approximately \( 8.85 \times 10^{-12} \text{ F/m} \) (farads per meter).

This constant is crucial when applying Gauss's Law, as it appears in the equation relating the electric field to the charge enclosed by the Gaussian surface. It essentially scales the relationship between charge and electric field in vacuum conditions.

Whenever calculating electric fields in nonconducting materials or discussing electrostatics in general, \( \varepsilon_0 \) provides essential reference values for predictions of how fields behave. In our exercise, it forms part of the denominator when calculating the electric field, representing the tendency of the vacuum to "allow" electric lines of force to pass through. Understanding its role is vital to grasping why certain configurations of charge yield their respective fields.