Problem 28 - Fundamentals of Physics

Question

A charge of uniform linear density \(2.0 \mathrm{nC} / \mathrm{m}\) is distributed along a long, thin, nonconducting rod. The rod is coaxial with a long conducting cylindrical shell (inner radius \(=5.0 \mathrm{~cm},\) outer radius \(=10 \mathrm{~cm}\) ). The net charge on the shell is zero. (a) What is the magnitude of the electric field \(15 \mathrm{~cm}\) from the axis of the shell? What is the surface charge density on the (b) inner and (c) outer surface of the shell?

Step-by-Step Solution

Answer

The electric field 15 cm from the axis is approximately 2.39 N/C. The inner surface charge density is -6.37 x 10^-9 C/m^2, and the outer is 6.37 x 10^-9 C/m^2.

Step 1: Understand the Problem

We have a linear charge density on a rod inside a conducting cylindrical shell. We need to calculate the electric field at a certain distance from the axis as well as the surface charge densities on the inner and outer surfaces of the shell.

Step 2: Calculate Electric Field Outside the Shell

The electric field due to a line of charge outside the shell can be found using Gauss's Law. The electric field at a distance \( r = 15 \mathrm{~cm} \) from the axis is given by:\[E = \frac{\lambda}{2 \pi \varepsilon_0 r}\] where \( \lambda = 2.0 \times 10^{-9} \mathrm{~C/m} \), \( r = 0.15 \mathrm{~m} \), and \( \varepsilon_0 \approx 8.85 \times 10^{-12} \mathrm{~C}^2/(\mathrm{N} \cdot \mathrm{m}^2) \). Substitute these values to find \( E \).

Step 3: Compute the Electric Field

Substituting the values into the formula:\[E = \frac{2.0 \times 10^{-9} \mathrm{~C/m}}{2 \pi (8.85 \times 10^{-12} \mathrm{~C}^2/(\mathrm{N} \cdot \mathrm{m}^2)) \times 0.15 \mathrm{~m}} \approx 2.39 \mathrm{~N/C}\]This is the magnitude of the electric field at a distance 15 cm from the axis of the shell.

Step 4: Surface Charge Density on Inner Surface

Because the net charge on the shell is zero, the surface charge density on the inner surface is equal and opposite to that of the linear charge density of the rod.\[ \sigma_{\text{inner}} = -\frac{\lambda}{2 \pi R_i} \]where \( \lambda = 2.0 \times 10^{-9} \mathrm{~C/m} \) and \( R_i = 0.05 \mathrm{~m} \). Calculating this gives.

Step 5: Compute Surface Charge Density on Inner Surface

\[\sigma_{\text{inner}} = -\frac{2.0 \times 10^{-9}}{2 \pi \times 0.05} \approx -6.37 \times 10^{-9} \mathrm{~C/m^2}\]This reflects the surface charge density on the inner surface of the shell.

Step 6: Surface Charge Density on Outer Surface

Since the net charge of the cylindrical shell is zero, the charge density on the outer surface equals that on the inner surface in magnitude but opposite in sign.\[ \sigma_{\text{outer}} = -\sigma_{\text{inner}} \]Thus, \( \sigma_{\text{outer}} = 6.37 \times 10^{-9} \mathrm{~C/m^2} \).

Key Concepts

Electric Field

An electric field represents a region around a charged object where forces are exerted on other charges.
In this exercise, we consider an electric field generated by a long, uniformly charged, nonconducting rod within a conductive cylindrical shell.
To find the electric field magnitude at a certain distance from the shell's axis, we utilize Gauss's Law. Gauss's Law provides a method to relate the electric field to the charge enclosed by a chosen surface. It states that the electric flux through a closed surface is proportional to the enclosed charge:\[\Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{ ext{enc}}}{\varepsilon_0}\]For a line of charge inside a conducting shell, the electric field at a distance \( r \) from the line charge is:\[E = \frac{\lambda}{2 \pi \varepsilon_0 r}\]Where:

• \( E \) is the electric field
• \( \lambda \) is the linear charge density
• \( \varepsilon_0 \) is the permittivity of free space
• \( r \) is the radial distance from the charge

In our exercise, calculating with \( \lambda = 2.0 \times 10^{-9} \mathrm{~C/m} \) and \( r = 0.15 \mathrm{~m} \), we find \( E \approx 2.39 \mathrm{~N/C} \). This result tells us the magnitude of the electric field at the specified distance.

Surface Charge Density

Surface charge density \( \sigma \) defines the amount of charge per unit area on a surface.
In this problem, it helps us understand how charges are distributed over the surfaces of the cylindrical shell.
The surface charge density on the inner surface of a conducting shell is key to maintaining the electric neutrality of the shell. Given the net charge of the shell is zero, and knowing the charge on the rod, the inner charge surface density is calculated using:\[\sigma_{\text{inner}} = -\frac{\lambda}{2 \pi R_i}\]Here, \( R_i \) is the inner radius of the shell.
By using \( \lambda = 2.0 \times 10^{-9} \mathrm{~C/m} \) and \( R_i = 0.05 \mathrm{~m} \), we find:\[\sigma_{\text{inner}} \approx -6.37 \times 10^{-9} \mathrm{~C/m^2}\]This negative value corresponds to the inward-facing surface of the inner shell. These negative charges counter the positive charges from the rod to neutralize the inner surface.

Linear Charge Density

Linear charge density \( \lambda \) is the charge per unit length along a line of charge. It describes how charge is spread along the rod.
For our rod with a constant linear charge density, its value is given as \( 2.0 \mathrm{nC/m} \), which means for every meter of the rod, \( 2.0 \times 10^{-9} \text{ C} \) of charge is present.
Understanding linear charge density is crucial because it influences how the electric field behaves around the rod.
With a uniformly charged rod, the electric field is evenly distributed, and we can use formulas that depend on linear charge density, such as when applying Gauss's Law.
This density also determines the amount of surface charge required to neutralize a conductor enclosing the charged rod, as seen with the surface charge density on the cylindrical shell in our exercise.