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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 distance 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 moves towards the center, accelerating and decelerating, reaching a speed of about 4.99 million m/s at the center.

Step by step solution

01

Understanding the System

The problem involves a uniformly charged ring with radius \( R = 15.0 \, \text{cm} \) and a total charge \( Q = +24.0 \, \text{nC} \). An electron, which carries a charge \( q_e = -1.6 \times 10^{-19} \, \text{C} \), is initially placed on the ring's axis at a distance \( z = 30.0 \, \text{cm} \). The electron is released from rest on the axis of the ring.
02

Calculating the Electric Field on the Axis of the Ring

The electric field \( E \) at a point on the axis of a charged ring is given by:\[ E = \frac{kQz}{(z^2 + R^2)^{3/2}} \]where \( k \) is the Coulomb's constant \( 8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2 \), \( Q \) is the charge of the ring, \( z \) is the distance from the center on the axis, and \( R \) is the radius of the ring.
03

Applying Newton's Second Law to Motion

The force \( F \) acting on the electron is given by \( F = q_e E \). Hence,\[ F = q_e \times \frac{kQz}{(z^2 + R^2)^{3/2}} \]Since the electron is released from rest at \( z = 30.0 \, \text{cm} \), it will accelerate towards the center of the ring under the influence of this force.
04

Assessment of Electron Motion (Qualitative)

As the force is attractive (since the electron is negatively charged and the ring is positively charged), the electron's motion will be directed towards the center of the ring. Thus, as it gets closer, the electric field and the exerted force decrease. The electron accelerates initially and decelerates as it moves towards the center of the ring, due to decreasing electric field strength.
05

Energy Conservation to Find Final Speed

Using the principle of energy conservation:\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]Initially, the electron is at rest, so its kinetic energy is zero, while its potential energy is given by:\[ U_i = -\frac{kQq_e}{(z^2 + R^2)^{1/2}} \]At the center of the ring, potential energy \( U_f = -\frac{kQq_e}{R} \), and we let \( v \) be the speed at the center of the ring:\[ \frac{1}{2}mv^2 = U_i - U_f \]Solve for \( v \) to find the speed of the electron.
06

Calculating Final Speed

Set up the equation:\[ \frac{1}{2}mv^2 = \frac{kQq_e}{R} - \frac{kQq_e}{(z^2 + R^2)^{1/2}} \]Solve for \( v \):\[ v = \sqrt{\frac{2kQq_e}{m}\left(\frac{1}{R} - \frac{1}{(z^2 + R^2)^{1/2}}\right)} \]Substitute the values:- \( k = 8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2 \)- \( Q = 24.0 \times 10^{-9} \, \text{C} \)- \( q_e = -1.6 \times 10^{-19} \, \text{C} \)- \( m = 9.11 \times 10^{-31} \, \text{kg} \) (mass of electron)- \( R = 0.15 \, \text{m} \)- \( z = 0.30 \, \text{m} \)Calculating gives \( v \approx 4.99 \times 10^6 \, \text{m/s} \).

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics that states energy cannot be created or destroyed, only transformed from one form to another. This concept is key in analyzing the motion of an electron in the field created by the charged ring. In this exercise, an electron is initially at rest at a certain point on the ring's axis.
At this point, the electron possesses potential energy due to its position in the electric field of the charged ring.
As it starts to move towards the ring's center due to the electric force, this potential energy is converted into kinetic energy, which increases as the electron speeds up.
  • Initially, all the energy is in the form of potential energy.
  • No kinetic energy is present because the electron starts from rest.
As the electron moves, the potential energy decreases, converted into kinetic energy up to the point where the electron reaches the center of the ring. At this location, the kinetic energy is at its maximum, and the potential energy is lesser compared to the starting point. Analyzing energy transformations helps to compute the electron's velocity at the ring's center using the energy conservation equation. All calculations assume that no energy is lost to the environment, which provides a simplified but powerful way to predict the electron's behavior.
Electric Force
Electric force is the force exerted on a charge due to the presence of other charges. In the case of the charged ring, it acts as a source of an electric field, which influences the electron. The force on the electron can be calculated using the electric field produced by the ring. The formula for the electric field on the axis of a charged ring is:
\[ E = \frac{kQz}{(z^2 + R^2)^{3/2}} \]
In this formula, **k** is the Coulomb's constant, **Q** is the charge of the ring, **z** is the axial distance from the center of the ring, and **R** is the ring's radius.
  • The electron experiences a force due to the electric field, calculated as \( F = q_e E \), where **q_e** is the charge of the electron.
  • This force is attractive because the electron is negatively charged while the ring is positively charged.
This attractive electric force causes the electron to accelerate towards the ring. As it moves closer, the electric field and the force decrease, resulting in reduced acceleration. Therefore, even as the electron continues to move towards the center, its acceleration diminishes over time.
Electron Motion
Understanding the concept of electron motion in an electric field requires examining the forces and resulting movement. The electron, influenced by the electric field of the charged ring, starts from a standstill and begins to move due to the attractive force exerted on it by the ring.
Initially, the electron accelerates as it is pulled towards the center of the ring. This movement is described by Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object times its acceleration \( F = ma \).
  • The force depends on its distance from the center, as we discussed in the electric force section.
  • Since the force decreases as the electron moves towards the ring, so does the acceleration.
The electron's velocity increases as it approaches the center, transforming electrical potential energy into kinetic energy. Upon reaching the center of the ring, the electron achieves its maximum speed, as all the potential energy that was available initially has now been converted to kinetic energy.
This motion concludes by helping to understand why the electron's speed is greatest at the ring's center, giving a practical insight into how charges move through electric fields.

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Most popular questions from this chapter

Nuclear Fusion in the Sun. The source of the sun's energy is a sequence of nuclear reactions that occur in its core. The first of these reactions involves the collision of two protons, which fuse together to form a heavier nucleus and release energy. For this process, called nuclear fusion, to occur, the two protons must first approach until their surfaces are essentially in contact. (a) Assume both protons are moving with the same speed and they collide head-on. If the radius of the proton is \(1.2 \times 10^{-15} \mathrm{m},\) what is the minimum speed that will allow fusion to occur? The charge distribution within a proton is spherically symmetric, so the electric field and potential outside a proton are the same as if it were a point charge. The mass of the proton is \(1.67 \times 10^{-27} \mathrm{kg} .\) (b) Another nuclear fusion reaction that occurs in the sun's core involves a collision between two helium nuclei, each of which has 2.99 times the mass of the proton, charge \(+2 e\) , and radius \(1.7 \times 10^{-15} \mathrm{m}\) . Assuming the same collision geometry as in part (a), what minimum speed is required for this fusion reaction to take place if the nuclei must approach a center-to-center distance of about \(3.5 \times 10^{-15} \mathrm{m}\) ? As for the proton, the charge of the helium nucleus is uniformly distributed throughout its volume. (c) In Section 18.3 it was shown that the average translational kinetic energy of a particle with mass \(m\) in a gas at absolute temperature \(T\) is \(\frac{3}{2} k T\) , where \(k\) is the Boltzmann constant (given in Appendix F). For two protons with kinetic energy equal to this avprage value to be able to undergo the process described in part (a), what absolute temperature is required? What absolute temperature is required for two average helium nuclei to be able to undergo the process described in part (b)? (At these temperatures, atoms are completely ionized, so nuclei and electrons move separately.) (d) The temperature in the sun's core is about 1.5 \(\times 10^{7}\) K. How does this compare to the temperatures calculated in part (c)? How can the reactions described in parts (a) and (b) occur at all in the interior of the sun? (Hint: See the discussion of the distribution of molecular speeds in Section \(18.5 . )\)

A small sphere with mass 1.50 g hangs by a thread between two parallel vertical plates 5.00 \(\mathrm{cm}\) apart (Fig. \(\mathrm{P} 23.62 ) .\) The plates are insulating and have uniform surface charge densities \(+\sigma\) and \(-\sigma .\) The charge on the sphere is \(q=8.90 \times 10^{-6} \mathrm{C} .\) What potential difference between the plates will cause the thread to assume an angle of \(30.0^{\circ}\) with the vertical?

A point charge \(q_{1}=+2.40 \mu \mathrm{C}\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu \mathrm{C}\) moves from the point \(x=0.150 \mathrm{m}, \quad y=0\) to the point \(x=0.250 \mathrm{m},\) \(y=0.250 \mathrm{m} .\) How much work is done by the electric force on \(q_{2} ?\)

Two stationary point charges \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of 50.0 \(\mathrm{cm} .\) An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 \(\mathrm{cm}\) from the \(+3.00\) -n \(\mathrm{charge}\)?

An infinitely long line of charge has linear charge density \(5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton \(1.67 \times 10^{-27} \mathrm{kg}\) , charge \(+1.60 \times 10^{-19} \mathrm{C} )\) is 18.0 \(\mathrm{cm}\) from the line and moving directly toward the line at \(1.50 \times 10^{3} \mathrm{m} / \mathrm{s}\) . (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?

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