Two thin rods of length L lie along the x-axis, one between x = (1)/(2) a and x = (1)/(2) a + L

step 1 In this question we have two thin rods of LL. They are placed at the position shown and then eac

Two thin rods of length L lie along the x-axis, one between...

Two thin rods of length $L$ lie along the $x$-axis, one between $x = \frac{1}{2} a$ and $x = \frac{1}{2} a + L$ and the other between $x = -\frac{1}{2} a$ and $x = -\frac{1}{2} a - L$. 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 $$F = {Q^2 \over 4\pi\epsilon_0 L^2} ln [ {(a + L)^2 \over a(a + 2L)} ]$$ (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 $(1 + z) = z - \frac{1}{2} z^2 + \frac{1}{3} z^3 - \cdot\cdot\cdot$, valid for $\mid z \mid\ll1$. Carry $all$ expansions to at least order $L^2/a^2.$) Interpret this result.

Two thin rods of length $L$ lie along the $x$-axis, one between $x = \frac{1}{2} a$ and $x = \frac{1}{2} a + L$ and the other between $x = -\frac{1}{2} a$ and $x = -\frac{1}{2} a - L$. 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

$$F = {Q^2 \over 4\pi\epsilon_0 L^2} ln [ {(a + L)^2 \over a(a + 2L)} ]$$

(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 $(1 + z) = z - \frac{1}{2} z^2 + \frac{1}{3} z^3 - \cdot\cdot\cdot$, valid for $\mid z \mid\ll1$. Carry $all$ expansions to at least order $L^2/a^2.$) Interpret this result.

Key Concepts

Electric Field from a Continuous Charge Distribution

This concept involves calculating the electric field produced by an extended object with a continuous distribution of charge. Instead of considering a point charge, the charge is spread out over a length, area, or volume, and the contributions of each infinitesimal piece of charge are summed (integrated) to determine the resulting electric field at a particular point.

Uniform Charge Distribution

A uniform charge distribution refers to a situation where charge is spread evenly over a specified region. In electrostatic problems, this assumption simplifies calculations because the charge density remains constant throughout the region, allowing for straightforward integration to determine quantities like the electric field and force.

Coulomb's Law and the Principle of Superposition

Coulomb's law quantifies the force between two point charges, and it serves as a building block for understanding interactions in systems with distributed charges. The principle of superposition states that the total electric field or force from a collection of charges is the vector sum of the fields or forces produced by each charge individually, which is fundamental in analyzing continuous charge configurations.

Integration in Electrostatics

Integration is a mathematical technique used to sum the contributions of infinitesimal charge elements in a continuous charge distribution. In the context of electrostatics, integration is applied to calculate the total electric field or potential by considering how each small element contributes to the overall behavior of the system.

Series Expansion and Approximation Techniques

Series expansion, such as the Taylor or Maclaurin series, provides a method to approximate complex functions by expanding them into simpler terms. In the context of electrostatics, these approximations are useful for studying limiting cases, such as when one length scale is much smaller than another, which helps in analyzing the behavior of force or field expressions in regimes where exact solutions may be cumbersome.

Limiting Cases in Electrostatic Forces

Analyzing limiting cases, such as when the separation between objects is much larger than their sizes, allows one to simplify the expressions for physical quantities. This approach aids in verifying that complex systems conform to well-known results (e.g., the inverse square law of Coulomb’s force) under certain conditions, providing both checks on the calculations and physical insight into the behavior of the system.

Transcript

00:01 In this question we have two thin rods of ll.

00:04 They are placed at the position shown and then each of them carry charge q uniformly distributed along the rod.

00:15 We want to find there are three parts in this question.

00:18 You need to find the electric field due to the left rod along the positive x -axis.

00:24 Then find the electric force due to that the rods exit on each other and show that the formula for the electric force reduced to the point charge formula when the distance between the rods are much larger than the length of the rod.

00:48 Okay, so in this question we are working with continuous charge distribution.

00:54 So we are going to cut the left rod.

00:59 Into very small pieces.

01:02 And at the same time, we also need to define the linear charge density.

01:07 Okay, of the rod.

01:10 Okay, so this is lambda, which is q over l.

01:13 Then this is our dq that we are going to consider.

01:19 Okay, and then for this dq, it exits, it generates an electric field at this point x.

01:27 Okay, that's pointing to the right.

01:30 Okay so the using the e equals to 1 over for pi xlophe x .l not dq over r square okay so so our r is measured from dq to the point x okay so i'm going to call this x prime okay so our dq is lambda the x this is q over l the x okay and then our r is x plus x prime okay so the x is fixed okay and then x prime is our integration variable okay okay so i'm going to put the dq as lambda the x prime and the x prime okay so we i'm going to put the dq as lambda the x prime and the x prime okay so we can set up our integral now, so our x prime, okay, goes from minus a over 2 minus l to minus a over 2.

02:55 This is our limits of integration.

03:07 Right, so now we can find the electric field due to the left rod along the positive x -axis.

03:14 So we'll integrate de, here we integrate 1 over.

03:23 For pi epsilon, q over l d x prime divide by x plus x prime square okay then the integration starts from minus a over 2 minus l to minus a over 2 okay so i'm going to pull out the constants q over for pi epsilon l okay minus a over 2 minus a over 2 minus a from minus a or two with minus l to minus a or two to two and then this is one over x plus x prime square x prime okay okay so we can integrate this directly okay so okay so this is just because we are integrating respect to x -prime and then the numerator is just x prime so the denominator is only has x -prime so the denominator is only has x -prime so so if you differentiate that, you get one.

04:34 So this means that you can actually obtain this expression.

04:45 Okay, immediately.

04:57 Yes.

04:59 Okay, one thing i forgot to mention that in this diagram, the x prime is negative, okay? so to be exact, okay, this should be x minus x prime, okay? because x prime is a negative number okay it's a negative number and so there is no negative sign over here okay you you'll obtain this yeah i hope this is clear okay x prime itself is negative okay so uh the r will be x minus x prime as mentioned in this box okay so then r will be a big number it's not plus okay yeah okay so when you do the substitution of the limits okay then you get one over x plus a or two minus one over x plus a over two plus l okay so you can choose to simplify or you can just leave it like that okay uh and both are fine okay so this is the answer or part a okay, then in part b, we want to find a force that the rod exits on each other.

06:29 So using, okay, so what happens is that in our situation...

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