/*! 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 53 A ring-shaped conductor with rad... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A ring-shaped conductor with radius \(a=2.50 \mathrm{cm}\) has a total positive charge \(Q=+0.125 \mathrm{nC}\) uniformly distributed around it, as shown in Fig. \(21.23 .\) The center of the ring is at the origin of coordinates \(O .\) (a) What is the electric field (magnitude and direction) at point \(P,\) which is on the \(x\) -axis at \(x=40.0 \mathrm{cm}\) ? (b) A point charge \(q=-2.50 \mu C\) is placed at the point \(P\) described in part (a). What are the magnitude and direction of the force exerted by the charge \(q\) on the ring?

Short Answer

Expert verified
(a) Electric field's magnitude is \(2.20 \times 10^3 \text{ N/C}\) along the positive \(x\)-axis. (b) The force magnitude is \(5.50 \text{ N}\) directed toward the negative \(x\)-axis.

Step by step solution

01

Understand the Configuration

We have a ring with radius \(a = 2.50 \text{ cm} = 0.025 \text{ m}\) carrying a total charge \(Q = 0.125 \text{ nC} = 0.125 \times 10^{-9} \text{ C}\). The ring is centered at the origin, and we want to find the electric field at point \(P\) on the \(x\)-axis at \(x = 40.0 \text{ cm} = 0.4 \text{ m}\).
02

Use the Electric Field Formula for a Ring

The electric field at a point along the axis of a ring of charge is given by the formula:\[E = \frac{kQx}{(x^2 + a^2)^{3/2}}\]where \(k = 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2\) is Coulomb's constant, \(x = 0.4 \text{ m}\) and \(a = 0.025 \text{ m}\).
03

Calculate the Electric Field at Point P

Substitute the known values into the equation:\[E = \frac{(8.99 \times 10^9) \times (0.125 \times 10^{-9}) \times 0.4}{(0.4^2 + 0.025^2)^{3/2}}\]Solve for \(E\) to get the magnitude of the electric field.
04

Evaluate the Expression

Step-by-step calculation gives:\[E = \frac{(8.99 \times 10^9) \times (0.125 \times 10^{-9}) \times 0.4}{(0.160625)^{3/2}}\]Continuing the calculation:\[E \approx \frac{0.44975}{0.2039} \approx 2.20 \times 10^3 \text{ N/C}\]
05

Determine the Electric Field Direction

Since the charge on the ring is positive and point \(P\) is along the positive \(x\)-axis, the electric field direction is along the positive \(x\)-axis.
06

Understand Force Between Charge and Ring

The point charge \(q = -2.50 \mu C = -2.50 \times 10^{-6} \text{ C}\) is placed at point \(P\). The force \(F\) exerted by the electric field on the charge is given by:\[ F = qE \]
07

Calculate the Force on the Charge

Substitute \(E = 2.20 \times 10^3 \text{ N/C}\) and \(q = -2.50 \times 10^{-6} \text{ C}\):\[ F = (-2.50 \times 10^{-6}) \times (2.20 \times 10^3) \approx -5.50 \text{ N} \]The negative sign indicates that the force is directed opposite to the electric field direction.
08

Interpret the Force Direction

Since the force is negative, it acts in the opposite direction to the electric field, i.e., toward the negative \(x\)-axis.

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.

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics, which describes the force between two point charges. According to this law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula is expressed as:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:
  • \(F\) is the electric force between the charges.
  • \(k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\) is Coulomb's constant.
  • \(q_1\) and \(q_2\) are the magnitudes of the charges.
  • \(r\) is the distance between the charges.
Understanding Coulomb's Law is essential because it forms the basis for calculating electric forces and fields between charges. In our exercise, while Coulomb's Law doesn't explicitly appear in the step-by-step solution, it underlies the principle of calculating forces involving electric fields and point charges.
Ring of Charge
The concept of a ring of charge involves a uniformly charged circular conductor. This is an important scenario in electrostatics as it allows us to understand the behavior of electric fields not just from point charges, but from distributed charges. For a ring of charge, however, directly applying Coulomb's Law is complex due to the continuous distribution of charge. Instead, we use the specific formula for the electric field created by a ring of charge.In the problem, the electric field \(E\) on the axis of a ring of charge is determined by:\[ E = \frac{kQx}{(x^2 + a^2)^{3/2}} \]where:
  • \(x\) is the distance from the center of the ring to the point where the field is calculated.
  • \(a\) is the radius of the ring.
  • \(Q\) is the total charge on the ring.
This approach simplifies the problem by taking into account symmetry, concentrating only on the net effect of the entire charge distribution along the central axial line.
Electric Force
Electric force is a fundamental aspect of electromagnetism. It’s the force of attraction or repulsion between charged objects. In this exercise, we focus on the electric force that acts upon a point charge in the presence of an electric field. The magnitude of this force is determined by the formula:\[ F = qE \]where:
  • \(F\) is the electric force.
  • \(q\) is the charge being subjected to the electric field.
  • \(E\) is the electric field strength.
In our example, a point charge \(q = -2.50 \mu C\) experiences a force due to the electric field at point \(P\). By using the formula \(F = qE\), where both \(q\) and \(E\) are known, we find the force magnitude. The negative sign of \(q\) indicates the force direction is opposite to the direction of the field. Thus, the concept of electric force not only quantifies interactions but also defines their directional characteristics.
Point Charge Interaction
A point charge is an idealized model utilized in physics to describe a charged object whose size is negligible compared to the distances involved in the problem. The interaction of a point charge with an electric field highlights the fundamental nature of electric forces. In this exercise, we are considering the interaction between the point charge placed at point \(P\) and the electric field produced by the ring of charge.These interactions can be understood through key concepts:
  • The electric field produced by the ring affects the point charge placed in it.
  • The point charge experiences a force due to this field, defined by \(F = qE\).
  • The direction of the force is influenced by the sign of the point charge in relation to the direction of the field.
The interaction is particularly interesting when charge is negative, as in this problem, leading the force to act in the opposite direction to the field. This reveals the symmetrical and predictable nature of electric forces and interactions, which are grounded in fundamental physical laws.

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

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.50 \(\mathrm{mm}\) away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.

CP An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 \(\mathrm{m}\) in the first 3.00\(\mu\) s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively,

What If We Were Not Neutral? A 75 -kg person holds out his arms so that his hands are 1.7 \(\mathrm{m}\) apart. Typically, a person's hand makes up about 1.0\(\%\) of his or her body weight. For round numbers, we shall assume that all the weight of each hand is due to the calcium in the bones, and we shall treat the hands as point charges. One mole of Ca contains \(40.18 \mathrm{g},\) and each atom has 20 protons and 20 electrons. Suppose that only 1.0\(\%\) of the positive charges in each hand were unbalanced by negative charge. (a) How many Ca atoms does each hand contain? (b) How many coulombs of unbalanced charge does each hand contain? (c) What force would the person's arms have to exert on his hands to prevent them from flying off? Does it seem likely that his arms are capable of exerting such a force?

Torque on a Dipole. An electric dipole with dipole moment \(\vec{p}\) is in a uniform electric field \(E\) . (a) Find the orientations of the dipole for which the torque on the dipole is zero. (b) Which of the orientations in part (a) is stable, and which is unstable? Hint: Consider a small displacement away from the equilibrium position and see what happens.)(c) Show that for the stable orientation in part (b), the dipole's own electric field tends to oppose the external field.

A straight, nonconducting plastic wire 8.50 \(\mathrm{cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 \(\mathrm{cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 \(\mathrm{cm}\) directly above its center.

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.