Problem 28 - Fundamentals of Physics

Question

Go 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 at 15 cm is \( 240 \; \text{N/C} \). The surface charge density on the inner surface is \( -11.5 \; \\mu \text{C/m}^2 \), and on the outer surface, it's \( 11.5 \; \\mu \text{C/m}^2 \).

Step 1: Understand the Configuration

We have a nonconducting rod with a linear charge density of \( \lambda = 2.0 \; \text{nC/m} \) distributed along a long axis. Coaxially around it is a conducting cylindrical shell with an inner radius of \( r_1 = 5.0 \; \text{cm} \) and an outer radius of \( r_2 = 10.0 \; \text{cm} \), and the net charge on the shell is zero.

Step 2: Electric Field Inside Conducting Shell

According to Gauss's law, inside the conductor, the electric field must be zero. This implies that for any point inside the shell, especially beyond the nonconducting rod but within the inner radius of the shell, \( E = 0 \). This ensures charge equilibrium in the conducting shell.

Step 3: Electric Field Outside the Shell

For a point outside the conducting shell, use Gauss's law. Any charge enclosed within our Gaussian surface at a distance \( r = 15 \text{ cm} \) would be solely from the nonconducting rod since the net charge on the shell is zero. The electric field \( E \) can be found using: \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]. Substitute \( \lambda = 2.0 \; \text{nC/m} = 2.0 \times 10^{-9} \; \text{C/m} \) and \( r = 0.15 \; \text{m} \).

Step 4: Calculate Electric Field Magnitude

Substitute into the formula: \[ E = \frac{2.0 \times 10^{-9}}{2\pi (8.85 \times 10^{-12}) (0.15)} \].Compute through the calculation to find \( E \).

Step 5: Surface Charge Density on Inner Surface

Due to electrostatic induction, the charge induced on the inner surface of the shell will be equal and opposite to that of the linear charge density on the rod. Let \( \sigma_{\text{inner}} \) represent this surface charge density. Given the net charge is zero, all induced charges on the inner surface compensate for the nonconducting rod's charge.

Step 6: Calculate Surface Charge Density on Inner Surface

The surface charge density depends on the charge per unit length and the area per unit length available: \[ \sigma_{\text{inner}} = -\frac{\lambda}{2\pi r_1} \].Substitute \( \lambda = 2.0 \times 10^{-9} \; \text{C/m} \) and \( r_1 = 0.05 \; \text{m} \) to find \( \sigma_{\text{inner}} \).

Step 7: Surface Charge Density on Outer Surface

Given the net charge on the shell is zero and no additional external electric fields are affecting it, the charge density on the outer shell will effectively neutralize the total charge induced on the inner surface, thus restoring equilibrium.

Step 8: Confirm Balance of Charges

For the outer surface, since the net charge is zero, the outer surface's charge will effectively counterbalance the charge induced on the inner surface. Thus, \( \sigma_{\text{outer}} = -\sigma_{\text{inner}} \). This ensures equilibrium without any additional force exerted by the field.

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle that relates the electric field emanating from a closed surface to the charge enclosed within that surface. It's mathematically represented as: \[ \Phi_E = \oint E \cdot dA = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where \( \Phi_E \) is the electric flux, \( E \) is the electric field, \( dA \) is the differential area element of the closed surface, \( Q_{\text{enc}} \) is the total enclosed charge, and \( \varepsilon_0 \) is the vacuum permittivity constant.
Gauss's Law simplifies many symmetry-based calculations in electrostatics. For example, when dealing with a coaxial configuration like a long, thin charged rod and a surrounding cylindrical shell, Gauss's Law helps to determine the electric field both inside and outside the shell. Inside a conductor, as per Gauss's Law, the electric field must be zero, while outside, it can be calculated based on the enclosed charge. In these scenarios, choosing an appropriate Gaussian surface like a cylinder coaxial with the rod allows us to easily apply conservation of electric flux.

Surface Charge Density

Surface charge density, denoted as \( \sigma \), refers to the amount of electric charge per unit area on a surface. It's given by the expression \( \sigma = \frac{Q}{A} \), where \( Q \) is the charge and \( A \) is the area over which this charge is distributed.
In electrostatics, when dealing with a conductive shell that encloses a charged nonconducting rod, charges will redistribute themselves across the surfaces of the shell. This redistribution happens due to electrostatic induction, a process where nearby charges influence the distribution of charges in a conductor.

• For the inner surface of the shell: The surface charge density, \( \sigma_{\text{inner}} \), equates to the linear charge density of the rod, but with an opposite sign, since all the induced charges on the inner surface must neutralize the rod's charge for the shell's net charge to remain zero.
• For the outer surface of the shell: The surface charge density, \( \sigma_{\text{outer}} \), will also ensure that the entire shell has a net charge of zero by mirroring and balancing any internal imbalances caused by the charges within.

This charge balancing maintains electrostatic equilibrium, which is key to understanding phenomena like shielding and charge induction on shells.

Conducting Shell

A conducting shell is a hollow conductor, often used in physics problems to explore concepts like charge induction, electrostatic equilibrium, and electric shielding. The conducting shell in our problem investigates the behavior of charge distribution and electric fields around it when it encloses another charged object.
A critical feature of conducting shells is their ability to distribute free charges uniformly across their surface, leading to an electric field of zero inside the shell material itself. This characteristic ensures:

• Any excess charge resides on the outer surface, creating an electrostatic equilibrium.
• Any electric field originating inside the shell, such as from a coaxial charged rod, causes redistribution of surface charges to counterbalance this field internally.

In our exercise, since the net charge of the shell is zero, all induced charges on the inner surface must precisely balance the rod’s linear charge density. Therefore, on the outer surface, equal but opposite charges will form, ensuring that the outside of the shell effectively shields any internal electric field to an external observer. This property is harnessed in designs like Faraday cages, which protect sensitive equipment from external electrostatic influences.