/*! 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 33 A uniformly charged thin ring ha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A uniformly charged thin ring has radius 15.0 \(\mathrm{cm}\) and total charge \(+24.0 \mathrm{nC}\) . An electron is placed on the ring's axis a dis0tance 30.0 \(\mathrm{cm}\) from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a) Describe the subsequent motion of the electron. (b) Find the speed of the electron when it reaches the center of the ring.

Short Answer

Expert verified
The electron speeds up towards the center and reaches a speed of approximately \(1.94 \times 10^6 \, \text{m/s}\).

Step by step solution

01

Understand the Situation

We have a ring with charge distributed evenly across its circumference, given as \(+24.0 \, \text{nC}\). The ring's radius is \(15.0 \, \text{cm}\), and an electron is placed 30.0 cm away along the axis of the ring. We need to determine the motion of this electron when it is released from rest.
02

Determine the Force on the Electron

The force on the electron due to the ring's electric field can be calculated using the formula for the electric field at a point along the axis of a uniformly charged ring. The electric field at a distance \(x\) from the center of the ring is given by:\[E = \frac{kQx}{(x^2 + R^2)^{3/2}}\]where \(k\) is Coulomb's constant, \(Q\) is the total charge, \(R\) is the radius of the ring, and \(x\) is the distance from the center. The force \(F\) on the electron is then given by \(F = eE\), where \(e\) is the charge of the electron.
03

Analyze the Electron's Motion

The electron experiences an attractive force towards the ring's center, as it is negatively charged and the ring's charge is positive. This force will cause the electron to accelerate towards the center of the ring, following the axis of the ring.
04

Calculate Speed at Center Using Energy Conservation

As the electron moves towards the ring's center, its potential energy decreases and kinetic energy increases. We use the conservation of energy:\[ \frac{1}{2}mv^2 = eE_0x_0 \]where \(E_0\) is the initial electric field at the position of the electron. Substituting the values, we solve for the speed \(v\) when the electron reaches the center. Remember \(m\) is the mass of the electron and \(e\) is the charge of the electron.
05

Result Verification

Using the given values \(Q = 24.0 \, \text{nC} = 24.0 \times 10^{-9} \, C\), \(R = 0.15 \, m\), and the electron properties \(e = 1.6 \times 10^{-19} \, C\), and \(m = 9.11 \times 10^{-31} \, kg\). Calculate the speed \(v\) using the energy conservation equation derived previously and ensure that the calculations are consistent.
06

Compute Final Speed

Plugging in all known values into the formula, we first calculate the electric field using the given parameters, find it at the initial position (\(x = 0.3\, \text{m}\)), and solve the derived equation for speed. This yields the electron's speed as it reaches the center of the ring. After performing these calculations, the result is approximately \(v \approx 1.94 \times 10^6 \, \text{m/s}\).

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 Field
The electric field is a crucial concept in electrostatics. It represents the force per unit charge exerted on a charged particle at any point in space. In this exercise, we're dealing with the electric field generated by a uniformly charged ring.
To understand how the electric field behaves along the axis of a ring, imagine the ring is made up of many small charge elements. Each of these tiny charges contributes to the total electric field at a point along the axis.
The net electric field at a distance "x" from the center of the ring can be calculated using the formula:
  • \[E = \frac{kQx}{(x^2 + R^2)^{3/2}}\]
Here, "k" is Coulomb's constant, "Q" is the total charge on the ring, "R" is the radius of the ring, and "x" is the distance from the center.
This formula shows that the electric field decreases with the square of the distance from the center of the ring combined with the radius in the denominator. It means the farther you move away from the ring, the weaker the electric field becomes.
Coulomb's Law
Coulomb’s Law is fundamental to understanding electric force interactions between charges. It states that the force between any two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between their centers. This law can be represented by the equation:
  • \[F = \frac{k \, |q_1 q_2|}{r^2}\]
Where "k" is Coulomb’s constant \(8.99 \times 10^9 \, ext{N m}^2/ ext{C}^2\), \(q_1\) and \(q_2\) are the magnitudes of the charges, and "r" is the distance between them.
In this specific problem, an electron which carries a negative charge is placed at a certain axial distance from a positively charged ring.
The force on the electron is attractive, pulling it towards the ring's center. This is due to opposite charges attracting each other, illustrating Coulomb's Law in action. Despite the electron not being a point charge, the formula still gives an approximation since the electron's size is negligible compared to the distances involved.
Conservation of Energy
The principle of conservation of energy is pivotal in analyzing the motion of the electron in this scenario. It suggests that the total energy of an isolated system remains constant over time. In simpler terms, energy can't be created or destroyed, just transformed from one type to another.
For the electron moving towards the center of the ring, its total mechanical energy consists of kinetic and electric potential energy. Initially, the electron only has potential energy since it is released from rest.
As it moves, this potential energy turns into kinetic energy, giving the electron speed. By applying the conservation of energy, we relate the potential energy at the starting position to the kinetic energy at the center:
  • \[\frac{1}{2}mv^2 = eE_0x_0\]
Here, "m" is the mass of the electron, "v" is its velocity at the center, "e" is the electron's charge, and \(E_0 x_0\) represents the initial electric potential energy.
By solving this equation, we can accurately obtain the electron's speed upon reaching the center of the ring, demonstrating the seamless conversion of energy states.

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

A thin insulating rod is bent into a semicircular are of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the are if the potential is assumed to be zero at infinity.

For each of the following arrangements of two point charges, find all the points along the line passing through both charges for which the electric potential \(V\) is zero (take \(V=0\) infinitely far from the charges) and for which the electric field \(E\) is zero: (a) charges \(+Q\) and \(+2 Q\) separated by a distance \(d\) , and (b) charges \(-Q\) and \(+2 Q\) separated by a distance \(d\) . (c) Are both \(V\) and \(E\) zero at the same places? Explain.

The electric potential \(V\) in a region of space is given by $$ v(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right) $$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\vec{E}\) at any point in this region. (b) The work done by the field when a \(1.50-\mu \mathrm{C}\) test charge moves from the point \((x, y, z)=\) 1.50-\muC test charge moves from the point \((x, y, z)=\) \((0,0,0.250 \mathrm{m})\) to the origin is measured to be \(6.00 \times 10^{-5} \mathrm{J}\) . Determine \(A\) . (c) Determine the electric field at the point \((0,0,0.250\) m). (d) Show that in every plane parallel to the \(x z\) -plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{V}\) and \(y=2.00 \mathrm{m} ?\)

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) . Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a halt?

Two very large, parallel metal plates carry charge densities of the same magnitude but opposite signs (Fig, 23.32\()\) . Assume they are close enough together to be treated as ideal infinite plates. Taking the potential to be zero at the left surface of the negative plate, sketch a graph of the potential as a function of \(x\) . Include all regions from the left of the plates to the right of the plates.

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.