Problem 13 - University Physics with Modern Physics

Question

Two very long uniform lines of charge are parallel and are separated by 0.300 m. Each line of charge has charge per unit length +5.20 \(\mu\)C/m. What magnitude of force does one line of charge exert on a 0.0500-m section of the other line of charge?

Step-by-Step Solution

Answer

The magnitude of the force is approximately 0.0257 N.

Step 1: Understand the Problem

We are given two parallel lines of charge, each with a linear charge density \( \lambda = +5.20 \, \mu C/m = 5.20 \times 10^{-6} \; C/m \). They are separated by a distance \( r = 0.300 \, m \). We need to find the force exerted by one line on a \( 0.0500 \, m \) section of the other line.

Step 2: Use Electric Field Formula for a Line of Charge

The electric field produced by an infinitely long line of charge at a perpendicular distance \( r \) is given by \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \), where \( \varepsilon_0 \) is the vacuum permittivity \( 8.85 \times 10^{-12} \, C^2/N \cdot m^2 \).

Step 3: Calculate the Electric Field

Substitute the values into the electric field formula: \( \lambda = 5.20 \times 10^{-6} \, C/m \) and \( r = 0.300 \, m \). Then, \( E = \frac{5.20 \times 10^{-6}}{2\pi(8.85 \times 10^{-12}) \times 0.300} \).

Step 4: Result of Electric Field Calculation

After performing the calculation, the electric field \( E \) is found to be approximately \( 9.87 \times 10^4 \, N/C \).

Step 5: Compute the Force on a Section of the Charge

The force on a charge in an electric field is given by \( F = E \cdot q \), where \( q = \lambda \times L \). Here, \( L = 0.0500 \, m \). Substitute \( \lambda \) into \( q \), \( q = 5.20 \times 10^{-6} \, C/m \times 0.0500 \, m = 2.60 \times 10^{-7} \, C \).

Step 6: Calculate the Force

Substitute \( E = 9.87 \times 10^4 \, N/C \) and \( q = 2.60 \times 10^{-7} \, C \) into the force formula: \( F = 9.87 \times 10^4 \, N/C \times 2.60 \times 10^{-7} \, C \).

Step 7: Result of Force Calculation

After calculation, the force \( F \) is approximately \( 2.57 \times 10^{-2} \, N \).

Key Concepts

Electric Field

An electric field is a region around a charged object where other charged objects experience a force. This force is caused due to the presence of the electric field created by the charge. The direction of the electric field is defined as the direction a positive test charge would move if placed in the field.
In the context of our problem, the electric field is produced by an infinite line of charge. For such a line, the field at a perpendicular distance \( r \) is given by the formula \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \). Here, \( \lambda \) is the linear charge density (charge per unit length) and \( \varepsilon_0 \) is the vacuum permittivity, an important constant that helps calculate electric forces in a vacuum. Vacuum permittivity is \( 8.85 \times 10^{-12} \, C^2/N \cdot m^2 \).

• To determine the field at a distance from a line of charge, plug the known values into the formula. In our case, \( \lambda = 5.20 \times 10^{-6} \, C/m \) and \( r = 0.300 \, m \).

After calculating, we found the electric field to be approximately \( 9.87 \times 10^4 \, N/C \), indicating the strength of the field at that distance from the charge line.

Coulomb's Law

Coulomb's Law describes the electrostatic interaction between two charged objects. It states that the force \( F \) between two stationary, point charges is directly proportional to their charge product and inversely proportional to the square of the distance \( r \) between them. The mathematical expression of Coulomb's Law is \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, N \cdot m^2/C^2 \).
In our case, instead of point charges, we have line charges. These require an understanding of how the linear charge density \( \lambda \) affects the force calculation.

• The electric field equation for a line of charge aligns with Coulomb's considerations, integrating the differential contributions of infinite charges along the line's length.
• Since we're dealing with a section of the line, the effective charge \( q \) acting over the distance is calculated as \( \lambda \times L \), where \( L = 0.0500 \, m \).

Using the previously calculated electric field, we can integrate the factors affecting the force using the term \( F = E \cdot q \). Our corrected force equation adapts to the line’s properties, providing insight into real-world charge interactions.

Linear Charge Density

Linear charge density \( \lambda \) is the amount of electric charge per unit length along a charged object. This concept is pivotal in electrostatics when dealing with objects like wires or lines instead of point charges.

• In simple terms, the linear charge density tells us how much charge is packed along a given segment of the line or rod.
• In the context of an infinite line of charge, as examined in our exercise, the linear charge density is given as \( \lambda = 5.20 \, \mu C/m \).

Understanding \( \lambda \) is crucial for calculating the electric field or force exerted by a charged line, as seen in the exercise. It directly influences both the electric field strength \( E \), calculated as \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \), and the force exerted \( F \), given by \( F = E \cdot q \), where \( q = \lambda \times L \).
By applying these concepts, students can predict and analyze the behavior of electrostatic forces in systems comprising continuous charges.