Problem 107 - University Physics with Modern Physics

Question

Two thin rods of length \(L\) le along the \(x\) -axis, one between \(x=a / 2\) and \(x=a / 2+L\) and the other between \(x=-a / 2\) and \(x=-a / 2-L .\) Each rod has positive charge \(Q\) distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive \(x\) -axis. (b) Show that the magnitude of the force that one rod exerts on the other is $$ F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right] $$ (c) Show that if \(a \gg L,\) the magnitude of this force reduces to \(F=Q^{2} / 4 \pi \epsilon_{0} a^{2} .\) (Hint: Use the expansion \(\ln (1+z)=z-\) \(z^{2} / 2+z^{3} / 3-\cdots,\) valid for \(|z| \ll 1 .\) Carry all expansions to at least order \(L^{2} / a^{2} .\) ) Interpret this result.

Step-by-Step Solution

Answer

The force between the rods simplifies to the force between point charges for large separation.

Step 1: Understanding the problem and finding the electric field

Given two rods with uniform charge \(Q\), we need to calculate the electric field from the second rod (lying from \(x=-a/2\) to \(-a/2-L\)) at a point along the positive \(x\)-axis. Treat a small element \(dx\) of charge \(dq = \frac{Q}{L} dx\) and use symmetry and properties of the electric field given by small elements to deduce the total field. The electric field due to a small charge segment is given by:\[ dE = \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{x^{2}} \]where \(x\) is the distance from the charge element to the point.

Step 2: Setting up the integral for the electric field

The differential charge element on the rod can be expressed in terms of \(dx\) as \(dq = \frac{Q}{L}dx\). For a point \(P\) on the positive \(x\)-axis, the distance to an element at \(x'\) on the rod is \(|x + x'|\), where \(x'\) is within \(-a/2 - L\) to \(-a/2\). Therefore, the differential electric field contribution is:\[ dE = \frac{\frac{Q}{L}dx}{4 \pi \epsilon_0 (x+x')^2} \]Integrate \(dE\) over the length of the rod from \(-a/2 - L\) to \(-a/2\) to find the total electric field at the point on the positive \(x\)-axis.

Step 3: Solve the integral for electric field

The integral to find the total electric field produced by the rod at a point on the positive \(x\)-axis is:\[ E = \int_{-a/2 - L}^{-a/2} \frac{Q}{4 \pi \epsilon_0 L (x+x')^2} dx' \]Compute this integral with respect to \(x'\) within the specified limits, using integral calculus. The result will describe the electric field dependent on the position \(x\) along the \(x\)-axis.

Step 4: Calculating the force between the rods

Using the field result from the second rod, integrate along the length of the first rod to find the force:\[ F = \int_{a/2}^{a/2+L} Q dE \]Substitute in the expression for \(dE\) from the total field expression, solve the integral to derive the force formula:\[ F = \frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2L)}\right] \]

Step 5: Simplifying the force for large \(a\) using a logarithmic expansion

Apply the expansion \(\ln(1+z) \approx z - \frac{z^2}{2} + \cdots\) for \(|z| \ll 1\). Here, \(L/a\) is very small compared to 1, to show:\[ F \approx \frac{Q^{2}}{4 \pi \epsilon_{0} a^{2}} \]This expansion reduces the complexity of the formula for the scenario where \(a \gg L\), confirming forces at large separations resemble point charges.

Step 6: Interpretation of the result

For \(a \gg L\), the two rods effectively interact like point charges. The force expression simplifies due to the logarithmic series expansion reflecting the diminishing interaction as the separation \(a\) gets much larger than the rod length \(L\). This is typical behavior for electric fields, leading to interaction forces diminishing with distance.

Key Concepts

Electric Field

The concept of electric fields is fundamental in understanding how charged objects interact over a distance. The electric field represents the force per unit charge exerted on a small positive test charge placed at a particular point in space. For a charge distribution, the electric field is a vector field around the charge that describes how the influence of the charge is distributed in the space around it.

The electric field of point charges can be calculated using Coulomb's law, and for continuous charge distributions, like our charged rod, we use calculus. By considering small segments of charge, we can compute the electric field by integrating the influence of each segment over the total length of the rod. This approach is crucial because it consolidates contributions from all segments, effectively building the entire field from its constituent parts.

• Contribution to the electric field by a charged rod involves slicing it into differential elements, where each introduces a small electric field at a point along the axis.
• The differential electric field due to a small charge element is calculated using a form of Coulomb's law modified for differential elements: \( dE = \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{r^2} \), where \( r \) is the distance to the point of interest.
• Summing up these contributions through integration yields the net electric field at a given point.

This process helps rationalize how charge distributions such as rods can set up electric fields in surrounding space, showing that fields can be non-uniform and directionally dependent.

Charged Rods

Charged rods provide an excellent real-world application of electrostatics. They represent extended charge distributions, unlike point charges, where the charge is smeared across a length. For this reason, they are more complex to handle because conventional point charge calculations cannot be applied as-is.

When two parallel charged rods are placed, the interaction between them depends on the charge distribution along each rod's length. The linear charge density, which is charge per unit length (\( \lambda = \frac{Q}{L} \)), is a critical parameter in these scenarios.

• For analysis, such rods are often divided into infinitesimally small charge elements. Each element behaves like a point charge contributing to the electric field or force.
• The mutual force between two charged rods can be determined by evaluating the influence of the electric field of one rod on the distributed charges of the other, using integration over the entire lengths.
• This model approximates realistic conditions where charged objects are not mere points but have spatial extent and complexity in charge positioning.

In exercises, students typically compute fields and forces between rods, giving a nuanced understanding that real-world charge interactions are length-dependent and involve integrating contributions across spatial domains.

Coulomb's Law

Coulomb's law is central to understanding electrostatic interactions, stating that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \]
This formula lays the groundwork for calculating forces in electrostatics and is extendable to continuous charge distributions by integrating the influences of infinitesimally small point charges over volume or length.

• When applied to systems of extended charges, like charged rods, Coulomb's law aids in determining the net force by considering the influence of each differential charge element along the distribution.
• The simplifying case for large separations, where distance \( a \) between charged rods is much larger than their lengths \( L \), uses expansions to approximate forces assuming point charge behavior.
• Logarithmic expansions, like \( \ln(1+z) \), assist in simplifying such complex interactions to resemble point charge scenarios, highlighting universality in electrostatic forces at large distances.

By mastering Coulomb's law, one gains insight into how charged particles exert forces on each other and how such forces diminish with increasing distance, a key principle in physics.