Problem 80 - University Physics with Modern Physics

Question

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside the surface of the wire. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.81 \lambda\) and radius \(6.5 r ?\)

Step-by-Step Solution

Answer

The magnitude of the electric field just outside the surface of the second wire is approximately 0.34 N/C.

Step 1: Find the electric field due to a line of charge

We can use the formula for the electric field due to a line of charge, which is given by: \(E = \frac{1}{4\pi\varepsilon_0}\frac{\lambda}{r}\) where E is the electric field, \(\lambda\) is the charge distribution, and r is the distance (here, it is the radius of the wire) from the line charge.

Step 2: Substitute the given values for the first wire

We are given the electric field \(E_1 = 2.73 \mathrm{~N} / \mathrm{C}\), charge distribution \(\lambda\), and radius \(r\). Plug these values into the formula from step 1: \(2.73 = \frac{1}{4\pi\varepsilon_0} \frac{\lambda}{r}\)

Step 3: Find the charge distribution of the second wire

The second wire has a charge distribution of \(0.81\lambda\). To find the magnitude of the electric field just outside the surface of this wire, we simply need to substitute this value in the formula from step 1.

Step 4: Find the radius of the second wire

The radius of the second wire is given as \(6.5r\). Again, we substitute this value in the formula.

Step 5: Solve for the magnitude of the electric field of the second wire

Plug the values of the charge distribution and radius for the second wire into the formula: \(E_2 = \frac{1}{4\pi\varepsilon_0}\frac{0.81\lambda}{6.5r}\) To find \(E_2\), we can rewrite the equation from step 2 as: \(\frac{1}{4\pi\varepsilon_0} = \frac{2.73}{\lambda/r}\) Now, plug this value into the equation for the second wire: \(E_2 = \frac{2.73}{\lambda/r} \frac{0.81\lambda}{6.5r}\) Simplify and solve for \(E_2\): \(E_2 = 2.73 \cdot \frac{0.81}{6.5} = \frac{2.73 \cdot 0.81}{6.5} \approx 0.34 \mathrm{~N} / \mathrm{C}\) So, the magnitude of the electric field just outside the surface of the second wire is approximately \(0.34 \mathrm{~N} / \mathrm{C}\).

Key Concepts

Charge Distribution

In the context of electricity and magnetism, charge distribution refers to how electric charge is spread or arranged over a given object or along a certain path. For this problem, we are dealing with linear charge distribution, denoted by \( \lambda \). Linear charge density \( \lambda \) is defined as the amount of charge per unit length along a line, such as a conducting wire or thread. This concept is essential when calculating electric fields around wires and rods.
Understanding charge distribution helps us apply Gauss's Law and other formulas effectively. When the charge distribution changes, as in this problem with one wire having \(0.81\lambda \), the resultant electric field will also change accordingly.

Conducting Wire

Conducting wires play a crucial role in electrical circuits and fields. They are designed to carry electric current efficiently. A key feature is that they are made of conductive material, allowing free movement of electrons. In a perfect conductor, charges distribute themselves uniformly on the surface to maintain electrostatic equilibrium.
When dealing with problems involving conducting wires, it's important to consider factors such as the wire's radius and the charge distribution along its length, as both directly affect the magnitude of the electric field produced. In this problem, a change in the radius from \(r\) to \(6.5r\) significantly influences the calculations and outcome.

Gauss's Law

Gauss's Law is a fundamental principle used to calculate electric fields, particularly in cases involving symmetry. The law states that the electric flux through a closed surface is directly proportional to the charge enclosed within that surface. Mathematically, it is expressed as: \[\Phi = \int E \, dA = \frac{Q_{enc}}{\varepsilon_0}\] where \(\Phi\) is electric flux, \(E\) is the electric field, \(dA\) is the differential area vector, \(Q_{enc}\) is the enclosed charge, and \(\varepsilon_0\) is the permittivity of free space.
Although Gauss's Law can simplify calculations, especially for symmetric charge distributions, it can be less intuitive. In this exercise, while Gauss's Law isn't explicitly used, understanding it supports the use of the electric field formula derived from it.

Line of Charge

A line of charge is an idealized model in physics that assumes charges are distributed along a straight line with a uniform charge density. This model simplifies the calculations of electric fields in many problems.
For any given point near a line of charge, the electric field strength relies on the charge density \(\lambda\) and the perpendicular distance \(r\) from that point to the line. The electric field due to a line of charge is given by \[ E = \frac{1}{2\pi\varepsilon_0} \frac{\lambda}{r} \] This formula demonstrates how the field decreases with increasing distance, a critical point in understanding electric fields around wires in this exercise. Notably, as charge density changes, so does the electric field, illustrating this core concept's importance in our calculations.