Problem 29 - Fundamentals of Physics

Question

Figure 23-42 is a section of a conducting rod of radius \(\quad R_{1}=1.30 \mathrm{~mm}\) and length \(L=11.00 \mathrm{~m}\) inside a thin-walled coaxial conducting cylindrical shell of radius \(R_{2}=10.0 R_{1}\) and the (same) length \(L .\) The net charge on the rod is \(Q_{1}=+3.40 \times 10^{-12} \mathrm{C} ;\) that on the shell is \(Q_{2}=-2.00 Q_{1} .\) What are the (a) magnitude \(E\) and (b) direction (radially inward or outward) of the electric field at radial distance \(r=2.00 R_{2} ?\) What are \((\mathrm{c}) E\) and \((\mathrm{d})\) the direction at \(r=5.00 R_{1} ?\) What is the charge on the (e) interior and (f) exterior surface of the shell?

Step-by-Step Solution

Answer

(a) E = 0 at r = 2.00 R2; (b) No direction at r = 2.00 R2; (c) E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\) at r = 5.00 R1; (d) Direction is outward; (e) Charge on interior surface = -Q1; (f) Charge on exterior surface = -Q1.

Step 1: Identify the Problem

We have a conducting rod and a coaxial conducting shell with given radii. Charges are distributed on these components. The task is to determine the electric field at given distances, and to evaluate the charge distribution on the shell.

Step 2: Calculate Electric Field at r=2.00 R2

Given that the charge on the shell is \(Q_2 = -2.00 Q_1\), the total charge in the system becomes zero when considering the charge inside the shell, leading to \(Q_{total} = Q_1 + Q_2 = 0\). Consequently, at any point beyond \(R_2\), the electric field is zero due to symmetrical charge cancellation.

Step 3: Determine Direction at r=2.00 R2

Since the electric field is zero at \(r = 2.00 R_2\), the direction is undefined or none. Electric fields only have direction when their magnitude is non-zero.

Step 4: Calculate Electric Field at r=5.00 R1

At \(r = 5.00 R_1\), which is inside the shell but outside the rod, we use Gauss's law. The linear charge density, \(\lambda\), is calculated as \(\lambda = \frac{Q_1}{L}\). The electric field at a distance \(r\) from the rod is given by \(E = \frac{\lambda}{2 \pi \varepsilon_0 r}\). Substitute \(r = 5.00 R_1\) to find \(E\).

Step 5: Determine Direction at r=5.00 R1

The electric field direction is determined by the sign of the charge on the rod. Since \(Q_1 = +3.40 \times 10^{-12} \, C\) is positive, the direction of the electric field will be radially outward from the rod.

Step 6: Find Charge on Interior Surface of the Shell

To find the charge on the inner surface of the shell, consider that the electric field inside the shell must be zero (due to conductor properties). Therefore, the inner surface of the shell must cancel the charge of the rod, i.e., \(Q_{inner} = -Q_1\).

Step 7: Find Charge on Exterior Surface of the Shell

The remaining charge on the shell will be on the outer surface. Given \(Q_{total shell} = Q_2 = -2 Q_1\), and the inner surface has \(Q_{inner} = -Q_1\), therefore, the exterior surface charge \(Q_{outer} = Q_2 - Q_{inner} = -2Q_1 - (-Q_1) = -Q_1\).

Key Concepts

Conducting Rod

The conducting rod is a cylindrical piece of material that can freely allow the movement of electric charges. This means that charges on the rod's surface can move until they have reached an equilibrium.
The charges spread evenly over the rod's surface due to the repulsive forces between like charges. In our problem, the conducting rod has a given radius and a fixed length, and it bears a net positive charge, denoted as \(Q_1 = +3.40 imes 10^{-12} \text{ C}\). This charge distribution affects how the electric field emanates from the rod, with the direction typically radially outward from the rod due to the positive charge nature.
For instance, at a radial distance of \(r = 5.00 R_1\), which is 5 times the radius of the rod, we calculate the electric field produced by the rod using the linear charge density and a key principle like Gauss's Law.

Coaxial Conducting Shell

A coaxial conducting shell is essentially a hollow cylindrical shell that surrounds another conductor, such as a rod in our exercise, sharing a central axis. This structure is frequently used in electrical systems to shield sensitive components from external electric fields by confining the electric field within the shell.
The coaxial shell in this problem has its own radius and bears a net negative charge given by \(Q_2 = -2.00 Q_1\). This negative charge works to neutralize or balance the positive charge on the rod, especially useful in calculating the electric field at any point outside the shell.
Importantly, coaxial shells can carry charges on both their inner and outer surfaces, altering potential applied fields, such as ensuring that fields are zero inside the conductor, as is the property of conductors to maintain equilibrium.

Charge Distribution

Charge distribution refers to how electric charge is spread across a conductor. It depends on the geometry of the object and the properties of the material. In our situation, the charge on the conducting rod influences the electric field it generates around itself.
For conductors, such as the rod and the shell, charges reside on the surface, impacting how the electric field within and outside the body is zero or non-zero. The conducting shell counteracts the rod's charge for the two to reach a point where their fields cancel each other out beyond their immediate vicinity, such as at any point outside the shell where the electric field was calculated to be zero.
Practically, understanding charge distribution assists in solving for charges on the surfaces of the shell—determining that the inner surface holds a charge that counteracts the rod \(-Q_1\), and the remaining charge resides on the outer surface.

Gauss's Law

Gauss's Law is a fundamental rule in electromagnetism, stating that the total electric flux through a closed surface is proportional to the enclosed charge. This can be mathematically expressed as \(\oint E \cdot dA = \frac{Q_{enclosed}}{\varepsilon_0}\). Here \(E\cdot dA\) denotes the electric field through a differential area \(dA\), and \(\varepsilon_0\) is the permittivity of free space.
In our exercise, Gauss's Law is used to determine the electric field at specific points outside the rod and shell arrangement. For instance, when calculating at \(r = 5.00 R_1\), only the charge within a Gaussian surface, which here is the charges of the rod, affects the field.
This law also helps clarify how charge distributions on conductors lead to zero internal electric fields, facilitating the examination of electric interactions between the conducting elements within the system. It is instrumental in confirming zero fields outside the shell due to complete charge cancellation between the rod and shell.