/*! 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 57 Point charges \(q_{1}=-4.5 \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by 3.1 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{N} \cdot \mathrm{m} ?\)

Short Answer

Expert verified
(a) Electric dipole moment is \( 1.395 \times 10^{-11} \, \mathrm{C} \cdot \mathrm{m} \), directed from \( q_1 \) to \( q_2 \). (b) Electric field magnitude is \( 8597.14 \, \mathrm{N/C} \).

Step by step solution

01

Understand the Given Data and Definitions

We have two point charges forming an electric dipole: \( q_1 = -4.5 \, \mathrm{nC} \) and \( q_2 = +4.5 \, \mathrm{nC} \), separated by a distance \( d = 3.1 \, \mathrm{mm} \). An electric dipole moment \( \mathbf{p} \) is defined as the product of one of the charges and the separation distance vector pointing from negative to positive charge.
02

Calculate the Magnitude of the Electric Dipole Moment

The magnitude of the electric dipole moment is \( p = q \, \times \, d \), where \( q = 4.5 \, \mathrm{nC} = 4.5 \times 10^{-9} \, \mathrm{C} \) and \( d = 3.1 \, \mathrm{mm} = 3.1 \times 10^{-3} \, \mathrm{m} \). Thus, \( p = 4.5 \times 10^{-9} \times 3.1 \times 10^{-3} = 1.395 \times 10^{-11} \, \mathrm{C} \cdot \mathrm{m} \).
03

Direction of the Electric Dipole Moment

The direction of the electric dipole moment vector \( \mathbf{p} \) is from the negative charge towards the positive charge, i.e., from \( q_1 \) to \( q_2 \).
04

Understand the Torque on an Electric Dipole in a Field

The torque \( \tau \) experienced by a dipole in an electric field \( \mathbf{E} \) is given by \( \tau = pE \sin\theta \), where \( \theta = 36.9^\circ \) is the angle between \( \mathbf{p} \) and \( \mathbf{E} \), and \( \tau = 7.2 \times 10^{-9} \mathrm{N} \cdot \mathrm{m} \).
05

Calculate the Magnitude of the Electric Field

Rearrange the torque equation \( \tau = pE \sin\theta \) to solve for \( E \): \( E = \frac{\tau}{p \sin \theta} \). Substitute \( \tau = 7.2 \times 10^{-9} \, \mathrm{N} \cdot \mathrm{m} \), \( p = 1.395 \times 10^{-11} \, \mathrm{C} \cdot \mathrm{m} \) and \( \sin(36.9^\circ) \approx 0.6 \) to get \( E = \frac{7.2 \times 10^{-9}}{1.395 \times 10^{-11} \times 0.6} \approx 8597.14 \, \mathrm{N/C} \).

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.

Electric Dipole Moment
An electric dipole consists of two equal and opposite charges separated by a small distance. The electric dipole moment, denoted as \( \mathbf{p} \), quantifies the strength and orientation of such a dipole system. It is a vector quantity pointing from the negative charge \( q_1 \) to the positive charge \( q_2 \).

In this particular exercise, we are given two charges \( q_1 = -4.5 \, \mathrm{nC} \) and \( q_2 = +4.5 \, \mathrm{nC} \), separated by a distance \( d = 3.1 \, \mathrm{mm} \). To calculate the dipole moment, the formula is:
\[ p = q \times d \]
where \( q \) is the magnitude of one of the charges and \( d \) is the separation distance. Here, \( q = 4.5 \times 10^{-9} \, \mathrm{C} \) and \( d = 3.1 \times 10^{-3} \, \mathrm{m} \). The calculated dipole moment is:
\[ p = 4.5 \times 10^{-9} \times 3.1 \times 10^{-3} = 1.395 \times 10^{-11} \, \mathrm{C} \cdot \mathrm{m} \]
The dipole moment has both magnitude and direction, oriented from the negative to positive charge.
Torque on a Dipole
When a dipole is placed in an external electric field, it experiences a torque. This torque tries to align the dipole moment with the electric field. The torque \( \tau \) exerted on the dipole is dependent on the magnitude of the dipole moment \( p \), the strength of the electric field \( E \), and the sine of the angle \( \theta \) between the dipole moment and the field:

\[ \tau = pE \sin(\theta) \]
In the exercise, the torque is given as \( 7.2 \times 10^{-9} \, \mathrm{N} \cdot \mathrm{m} \), and the angle \( \theta = 36.9^\circ \). This angle measures the initial misalignment between the dipole and the electric field.

Understanding the role of \( \sin(\theta) \) is crucial. It indicates that the torque is zero when the dipole is aligned completely parallel or anti-parallel with the electric field (since \( \sin(0^\circ) = \sin(180^\circ) = 0 \)). Maximum torque is felt when \( \theta = 90^\circ \). Here, the actual calculation revolves around resolving the equation to find the unknown electric field.
Uniform Electric Field
A uniform electric field is one where the field strength and direction are constant at every point in the area it occupies. This means that if we place a charge anywhere within the field, the force it experiences will always be the same magnitude and direction.

In the case of our exercise, such a field affects the dipole. We find the magnitude of this uniform electric field \( E \) using the torque experienced by the dipole:
\[ E = \frac{\tau}{p \sin(\theta)} \] Here, \( \tau = 7.2 \times 10^{-9} \, \mathrm{N} \cdot \mathrm{m} \), \( p = 1.395 \times 10^{-11} \, \mathrm{C} \cdot \mathrm{m} \), and \( \sin(36.9^\circ) \approx 0.6 \). Substituting these values gives:
\[ E \approx \frac{7.2 \times 10^{-9}}{1.395 \times 10^{-11} \times 0.6} \approx 8597.14 \mathrm{N/C} \]
With this calculation, we determine the field's strength, confirming its uniform nature. A consistent electric field like this simplifies many calculations involving electric forces and torques.

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

Two positive point charges \(q\) are placed on the \(x\) -axis, one al \(x=a\) and one at \(x=-a\) . (a) Find the magnitude and direction of the electric field at \(x=0 .\) (b) Derive an expression for the electric field at points on the \(x\) -axis. Use your result to graph the \(x\) -component of the electric field as a function of \(x,\) for values of \(x\) between \(-4 a\) and \(+4 a .\)

Two point charges are placed on the \(x\) -axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m} .\) What are the magnitude and direction of the total force exerted by these two charges on a negative point charge \(q_{3}=-6.00 \mathrm{nC}\) that is placed at the origin?

CP Two tiny spheres of mass 6.80 \(\mathrm{mg}\) carry charges of equal magnitude, \(72.0 \mathrm{nC},\) but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 \(\mathrm{m}\) . When a horizontal uniform electric field \(E\) that is directed to the left is turned on, the spheres hang at rest with the angle \(\theta\) between the strings equal to \(50.0^{\circ}(\) Fig. .21 .82\()\) . (a) Which ball (the one on the right or the one on the left) has positive charge? (b) What is the magnitude \(E\) of the field?

Three point charges are arranged along the \(x\) -axis. Charge \(q_{1}=+3.00 \mu C\) is at the origin, and charge \(q_{2}=-5.00 \mu C\) is at \(x=0.200 \mathrm{m} .\) Charge \(q_{3}=-8.00 \mu \mathrm{C} .\) Where is \(q_{3}\) located if the net force on \(q_{1}\) is 7.00 \(\mathrm{N}\) in the \(-x\) -direction?

CP A small sphere with mass 9.00\(\mu g\) and charge \(-4.30 \mu C\) is moving in a circular orbit around a stationary sphere that has charge \(+7.50 \mu \mathrm{C}\) . If the speed of the small sphere is \(5.90 \times 10^{3} \mathrm{m} / \mathrm{s},\) what is the radius of its orbit? Treat the spheres as point charges and ignore gravity.

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.