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

Name: Problem 97, Chapter 21: University Physics with Modern Physics
Uploaded: 2021-05-02T21:07:05-08:00
Duration: 11 min 44 s
Channel: Linda Winkler

Question

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 $F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right]$ (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 $\ln (1+z)=$ $z-\frac{1}{2} z^{2}+\frac{1}{3} z^{3}-\cdots,$ valid for $|z| \ll 1 .$ Carry all expansions to at least order $\left.L^{2} / a^{2} .\right)$ Interpret this result. Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be $+(30 \pm 5)$ pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain).

   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 $F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right]$ (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 $\ln (1+z)=$ $z-\frac{1}{2} z^{2}+\frac{1}{3} z^{3}-\cdots,$ valid for $|z| \ll 1 .$ Carry all expansions to at least order $\left.L^{2} / a^{2} .\right)$ Interpret this result. Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be $+(30 \pm 5)$ pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain).

University Physics with Modern Physics

Roger A. Freedman,… 15th Edition

Chapter 21, Problem 97

Instant Answer

Solved by Expert Linda Winkler

05/02/2021

Step 1

We consider a small element of charge dq at a distance x' from the origin on the second rod. The electric field dE at a point x due to this element is given by Coulomb's law: \[dE = \frac{1}{4\pi\epsilon_{0}} \frac{dq}{(x-x')^{2}}\] where dq = λdx' and λ is the Show more…

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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 $F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right]$ (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 $\ln (1+z)=$ $z-\frac{1}{2} z^{2}+\frac{1}{3} z^{3}-\cdots,$ valid for $|z| \ll 1 .$ Carry all expansions to at least order $\left.L^{2} / a^{2} .\right)$ Interpret this result. Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be $+(30 \pm 5)$ pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain).

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Key Concepts

Electrostatic Induction and Charge Redistribution

This concept pertains to the phenomenon where a charged object induces a redistribution of charges in a nearby neutral object, resulting in a net separation of charge or even a temporary net charge. Understanding induction is crucial when interpreting interactions where approaching charged bodies cause measurable effects in otherwise neutral systems.

Continuous Charge Distribution

This concept involves understanding how charge may be spread over a region rather than concentrated at a point. In such cases, the contribution from each infinitesimal element of charge must be summed—typically by integration—to obtain the overall electric field or potential generated by the object.

Electric Field Calculation from Extended Bodies

Calculating the electric field due to an extended object, such as a rod, uses the notion of superposition and integration of the contributions from each differential charge element. This process requires establishing the geometry of the distribution and integrating the differential field expressions over the charged object’s volume or length.

Application of Coulomb’s Law Through Integration

Coulomb’s law, which describes the force between two point charges, can be extended to continuous charge distributions by integrating the infinitesimal contributions of force from each small element of charge. This approach bridges the gap between simple point charge interactions and interactions between extended bodies.

Series Expansion Approximations

When dealing with small parameters, series expansions such as the Taylor or logarithmic expansion can be used to simplify otherwise complex functions. This technique allows physicists to extract approximate solutions in limiting cases, making it easier to interpret the physical behavior in regimes where certain dimensions are much smaller compared to others.

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Transcript

00:01 In this problem, we're going to find the force of interaction between two charge rods, and they both have the same charge per unit length on them, which we'll call lambda, and we'll see how that comes about.

00:18 But we're basically going to show how you can break up a charge distribution into small differential elements and use those small elements with integration to find the force between the charge.

00:32 Those two.

00:34 And the first step is to actually figure out the electric field at some point on the right.

00:46 So here's an observation point.

00:49 Find out the electric field at that point, p, due to the black rod.

00:58 That's over to the left of the origin.

01:01 So the way to do that is to imagine that that black rod is broken up into little tiny chunks of charge and actually use kulom's law to determine the net effect of all those charges.

01:19 So we'll kind of write that down.

01:22 And we have a one over a distance squared in there, or one over r squared.

01:30 And just to see what one of those little chunks of charge looks like, it's inside that rod.

01:36 But we know that electric fields superimpose.

01:41 So if we could write dq as some charge density times a spatial component, we could then integrate over the space of the object.

01:58 And so our spatial component is simply along the x direction.

02:03 And we have a linear charge density.

02:06 Now, r is the distance of separation.

02:09 Between our observation point minus our source point, which is inside that rod.

02:23 And we can see that we can easily write this as x minus x prime.

02:30 It's the separation between point p, which is fixed at position x, and the variable x prime that we are going to be integrating over.

02:42 So to find the electric field, simply integrate all the little contributions from all the little charges.

02:56 Yeah, there's a constant out front.

03:00 And then we want to actually integrate over that black rod.

03:03 So from x prime equals minus 1⁄2a, minus l, all the way to minus 1⁄2a.

03:18 And we have inside of that integral, of course, a dx prime, and then the x minus x prime squared.

03:34 Okay, and just a reminder that the x prime is the position of the little chunk of charge, and we want to find all the contributions from all those little chunks of charge.

03:45 And that's a fairly straightforward integral.

03:48 I won't write all the details down, but we find out that the total electric field due to the black rod is basically a 1 over x minus x prime type of deal.

04:04 We put in the limits of integration, and we have x plus 1 half a minus 1 over x plus 1⁄2a plus l.

04:23 So that's the first step.

04:27 And then the second step to find the force of interaction is to do this breakup business one more time.

04:36 But here we are going to break up the interactive force using the relationship that force is equal to q times electric field.

04:50 And i have to point out that now what we are going to break up is the charges in the blue rod, and we're going to integrate their contributions to the force of the blue rod.

05:13 So again, i'm probably not going to show you all the details of this integral, but we'll at least write it down.

05:30 And what we want to put in there is we're integrating over the blue rod, but we're using the electric field that we just found...

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