/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 (III) Suppose a dipole \(\vec { ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(III) Suppose a dipole \(\vec { \mathbf { p } }\) is placed in a nonuniform electric field \(\vec { \mathbf { E } } = E \hat { \mathbf { i } }\) that points along the \(x\) axis. If \(\vec { \mathbf { E } }\) depends only on \(\mathbf { x } ,\) show that the net force on the dipole is \(\vec { \mathbf { F } } = \left( \vec { \mathbf { p } } \cdot \frac { d \vec { \mathbf { E } } } { d x } \right) \hat { \mathbf { i } }\) where \(d \vec { \mathbf { E } } / d x\) is the gradient of the field in the \(x\) direction.

Short Answer

Expert verified
The net force on the dipole is \(\vec{\mathbf{F}} = \left( \vec{\mathbf{p}} \cdot \frac{d\vec{\mathbf{E}}}{dx} \right) \hat{\mathbf{i}}\).

Step by step solution

01

Understanding the Dipole and Electric Field Components

A dipole consists of two point charges of equal magnitude but opposite sign separated by a small distance. When placed in an electric field, each charge experiences a force due to the field. The electric field provided is \(\vec{\mathbf{E}} = E \hat{\mathbf{i}}\), indicating that it is uniform in direction (along \(\hat{\mathbf{i}}\)) but non-uniform in magnitude, as its value depends on the position \(x\).
02

Expression for Force on a Single Charge

The force on a single charge \(q\) in an electric field \(\vec{\mathbf{E}}\) is given by \(\vec{\mathbf{F}} = q \vec{\mathbf{E}}\). For an infinitesimal element of the dipole with charge \(q\) and separation \(dx\) from the center, the force would be \(q E(x+dx)\) and \(-q E(x)\).
03

Approximating the Electric Field at Displaced Charges

For the charge \(+q\) placed at \(x + a/2\) and charge \(-q\) at \(x - a/2\), we approximate: \(E(x + a/2) \approx E(x) + \frac{a}{2} \frac{dE}{dx}\) and \(E(x - a/2) \approx E(x) - \frac{a}{2} \frac{dE}{dx}\).
04

Calculating the Forces on Each Component of the Dipole

The force on \(+q\) will be \(q (E(x) + \frac{a}{2} \frac{dE}{dx})\), and the force on \(-q\) will be \(-q (E(x) - \frac{a}{2} \frac{dE}{dx})\).
05

Finding the Net Force on the Dipole

To find the net force on the dipole, sum the forces on the two charges: \(q \left(E(x) + \frac{a}{2} \frac{dE}{dx}\right) - q \left(E(x) - \frac{a}{2} \frac{dE}{dx}\right)\).
06

Simplifying the Net Force Expression

The terms \(qE(x)\) cancel out, leaving \(q \frac{a}{2} \frac{dE}{dx} + q \frac{a}{2} \frac{dE}{dx} = qa \frac{dE}{dx}\). Recognizing that the dipole moment \(\vec{\mathbf{p}}\) is given by \(qa\), the net force becomes \(\vec{\mathbf{F}} = \vec{\mathbf{p}} \cdot \frac{d\vec{\mathbf{E}}}{dx} \hat{\mathbf{i}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nonuniform Electric Field
An electric field is a region around a charged particle or object where a force is exerted on other charged particles. When this field is nonuniform, its intensity changes at different positions within the field. In the exercise, we encounter a nonuniform electric field denoted by \(\vec{\mathbf{E}} = E \hat{\mathbf{i}}\), which implies that although the direction of the field is consistent along the \(x\) axis, its magnitude depends on the specific location \(x\).
This dependency can be represented mathematically by the gradient \(\frac{d\vec{\mathbf{E}}}{dx}\), which measures how the strength of the electric field changes as you move along the \(x\) axis.
  • Nonuniform fields are essential in understanding how forces vary across different regions.
  • They play a crucial role in various physics applications, such as capacitors and electrical circuits.
Recognizing the nonuniform nature is critical for analyzing the behavior of dipoles within such fields.
Dipole Moment
A dipole consists of two charges that are equal in magnitude but opposite in sign, separated by a small distance. The dipole moment, represented as \(\vec{\mathbf{p}}\), is a vector quantity that indicates the strength and orientation of a dipole.
It is calculated as \(\vec{\mathbf{p}} = q \cdot a\), where \(q\) is the charge and \(a\) is the separation distance between the charges, with the direction of the dipole moment pointing from the negative to the positive charge.
  • The dipole moment is a measure of the system's overall charge distribution.
  • It plays a significant role in determining how the dipole interacts with an external electric field.
This concept helps us understand how dipoles are oriented and how they are affected when placed in electric fields.
Net Force on Dipole
The net force on a dipole in an electric field arises from the different forces experienced by each charge due to their positions. In a nonuniform electric field, these forces are unequal, resulting in a net force that acts on the dipole.
From the exercise, the forces on the two charges of the dipole are calculated, and their difference leads us to the net force expression. The electric force on each charge depends on their respective electric field strengths, as given by \(q E(x + \frac{a}{2})\) for the positive charge and \(-q E(x - \frac{a}{2})\) for the negative charge.
After canceling certain terms and simplifying, we find that the net force on the dipole is determined by \(\vec{\mathbf{F}} = \vec{\mathbf{p}} \cdot \frac{d\vec{\mathbf{E}}}{dx} \hat{\mathbf{i}}\).
  • This equation highlights the influence of the dipole moment and the nonuniform field on the net force.
  • Understanding this concept is vital for analyzing how electric dipoles behave in varying electric fields.
The net force leads to movement of the dipole in the direction where the field is increasing, showcasing how dipoles align according to the field gradient.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(II) An electron has an initial velocity \(\vec { \mathbf { v } } _ { 0 } = \left( 8.0 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } \right) \hat { \mathbf { j } }\) . It enters a region where \(\quad \vec { \mathbf { E } } = ( 2.0 \hat { \mathbf { i } } + 8.0 \hat { \mathbf { j } } ) \times 10 ^ { 4 } \mathrm { N } / \mathrm { C }\) (a) Determine the vector acceleration of the electron as a function of time. \(( b )\) At what angle \(\theta\) is it moving (relative to its initial direction) at \(t = 1.0 \mathrm { ns } ?\)

(III) A thin ring-shaped object of radius \(a\) contains a total charge \(Q\) uniformly distributed over its length. The electric field at a point on its axis a distance \(x\) from its center is given in Example 9 of "Electric Charge and Electric Field" as \(E = \frac { 1 } { 4 \pi \epsilon _ { 0 } } \frac { Q x } { \left( x ^ { 2 } + a ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\) (a) Take the derivative to find where on the \(x\) axis \(( x > 0 ) E _ { x }\) is a maximum. Assume \(Q = 6.00 \mu \mathrm { C }\) and \(a = 10.0 \mathrm { cm } .\) (b) Calculate the electric field for \(x = 0\) to \(x = + 12.0 \mathrm { cmin }\) steps of \(0.1 \mathrm { cm } ,\) and make a graph of the electric field. Does the maximum of the graph coincide with the maximum of the electric field you obtained analytically? Also, calculate and graph the electric field \(( c )\) due to the ring, and \(( d )\) due to a point charge \(Q = 6.00 \mu C\) at the center of the ring. Make a single graph, from \(x = 0\) (or \(x = 1.0 \mathrm { cm }\) ) out to \(x = 50.0 \mathrm { cm }\) in 1.0\(\mathrm { cm }\) steps, with two curves of the electric fields, and show that both fields converge at large distances from the center. (e) At what distance does the electric field of the ring differ from that of the point charge by 10\(\% ?\)

(III) Suppose a dipole \(\overrightarrow{\mathbf{p}}\) is placed in a nonuniform electric \(\overrightarrow{\mathbf{E}}=E \hat{\mathbf{i}}\) that points along the \(x\) axis. If \(\overrightarrow{\mathbf{E}}\) depends only on \(x\), show that the net force on the dipole is $$ \overrightarrow{\mathbf{F}}=\left(\overrightarrow{\mathbf{p}} \cdot \frac{d \overrightarrow{\mathbf{E}}}{d x}\right) \hat{\mathbf{i}} $$ where \(d \overrightarrow{\mathbf{E}} / d x\) is the gradient of the field in the \(x\) direction.

(I) A downward electric force of 8.4\(\mathrm { N }\) is exerted on a \(- 8.8 \mu \mathrm { C }\) charge. What are the magnitude and direction of the electric field at the position of this charge?

(II) Compare the electric force holding the electron in orbit \(\left( r = 0.53 \times 10 ^ { - 10 } \mathrm { m } \right)\) around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton. What is the ratio of these two forces?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.