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Chapter 23: Problem 27

A uniformly charged, thin ring has radius 15.0 cm and total charge \(+\)24.0 nC. An electron is placed on the ring's axis a distance 30.0 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 accelerates towards the center and gains speed; its speed is determined by converting initial potential energy into kinetic energy.

Step by step solution

01

Understanding the Problem

We have a ring with radius 15.0 cm and charge +24.0 nC. An electron is placed on the axis, 30.0 cm away, and is released. We'll first describe how the electron moves and then find its speed when it reaches the center.
02

Electron's Motion Description

The electron is negatively charged, and the ring is positively charged. Thus, the electron is attracted towards the center of the ring. It will accelerate towards the ring because of the electrostatic force and travel down the axis to the center.
03

Equipotential Surface Analysis

Understand that on the axis of a charged ring, there is an electric potential resulting from the charges uniformly distributed along the ring. The electron, being on the axis, will be subjected to this potential, hence will convert its potential energy into kinetic energy as it moves toward the center.
04

Calculation of Electric Potential and Energy Conversion

The potential energy of the electron at the start can be calculated using the electric potential due to the ring at a point on its axis. Given the ring's charge \( Q = 24.0 \, \text{nC} = 24.0 \times 10^{-9} \, \text{C} \), radius \( a = 0.15 \, \text{m} \) and distance \( x = 0.30 \, \text{m} \), the electric potential \( V \) is: \[ V = \frac{kQ}{\sqrt{a^2 + x^2}} \]Where \( k = 8.99 \times 10^9 \, \text{Nm}^2\/\text{C}^2 \). Calculate \( V \) and then the initial potential energy \( U_i = -eV \) for an electron with charge \( e = -1.6 \times 10^{-19} \, \text{C} \).
05

Kinetic Energy at the Center

When the electron reaches the center of the ring, its potential energy becomes zero. Therefore, the initial potential energy converts into kinetic energy, given as:\[ K_e = U_i = \frac{1}{2} mv^2 \]Where \( m = 9.11 \times 10^{-31} \text{kg} \) is the mass of the electron, and \( v \) is the speed we need to find. Rearrange to find \( v \).
06

Calculation of Speed

Set the initial potential energy equal to the kinetic energy: \[ -eV = \frac{1}{2} mv^2 \]Solve for \( v \) using the calculated value of \( V \) from Step 4 and plug into the equation for kinetic energy to find \( v \):\[ v = \sqrt{\frac{2(-eV)}{m}} \]Compute this to determine the electron's speed when it reaches the center.

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Key Concepts

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

Electrostatics
Electrostatics is the study of electric charges at rest, and it plays a crucial role in understanding how electric forces act over distances. In the scenario involving the charged ring and the electron, the principles of electrostatics help us predict the motion of the electron.
The charged ring creates an electric field in the space surrounding it due to its charge distribution. This field behaves like a map showing the direction and magnitude of the force experienced by a test charge, like our electron, placed in its vicinity.
  • Static charges create fields that influence other charges.
  • These fields in turn exert electrostatic forces, attracting or repelling charges around them.
The electron in this exercise is initially stationary, but the electrostatic forces come into play once released, guiding it towards the center of the charged ring.
Kinetic Energy
Kinetic energy refers to the energy that a body or particle possesses due to its motion. In this exercise, as the electron moves towards the center of the ring, it converts potential energy into kinetic energy.
The conservation of energy principle suggests that the total energy remains constant. Initially, the electron has maximum potential energy owing to its position relative to the electric potential of the ring. As the electron accelerates towards the center:
  • The potential energy decreases because the electron is moving to a lower electric potential.
  • This decrease in potential energy results in a corresponding increase in kinetic energy.
At the center of the ring, the electron鈥檚 potential energy is effectively zero, and all initial potential energy is converted into kinetic energy, allowing us to calculate its speed using the kinetic energy formula.
Electrostatic Force
The electrostatic force is the force exerted by a charged body on another due to their electric charges. It's one of the fundamental forces in physics, described by Coulomb's law. In this problem, this force is responsible for the acceleration of the electron towards the center of the ring.
Coulomb's Law tells us that the magnitude of the electrostatic force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between their centers. For our exercise:
  • The electron is negatively charged and the ring is positively charged.
  • The electrostatic force is attractive, working to pull the electron towards the ring.
As the distance decreases, the force on the electron increases, leading to greater acceleration towards the center.
Equipotential Surface
An equipotential surface is a surface where every point has the same electric potential. In this problem, because of the symmetry of the charged ring, an equipotential surface can be visualized along the axis of the ring where the electron lies.
When the electron moves along the axis, it travels through different equipotential levels, causing changes in its potential energy. However, while in complete contact with an equipotential surface, the electron's potential energy remains constant:
  • Movement perpendicular to an equipotential surface does not change the potential energy.
  • Potential energy changes only when moving through different equipotential surfaces.
This concept is crucial for understanding how the potential energy of the electron is converted into kinetic energy as it approaches the center of the ring.

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Most popular questions from this chapter

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is given by $$V(x) = Cx^{{4}/{3}}$$ where \(x\) is the distance from the cathode and \(C\) is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of \(C\). (b) Obtain a formula for the electric field between the electrodes as a function of \(x\). (c) Determine the force on an electron when the electron is halfway between the electrodes.

A positive point charge \(q_1 = +5.00 \times 10^{-4}\) C is held at a fixed position. A small object with mass 4.00 \(\times 10^{-3}\) kg and charge \(q_2 = -3.00 \times 10^{-4}\) C is projected directly at \(q_1\) . Ignore gravity. When \(q_2\) is 0.400 m away, its speed is 800 m\(/\)s. What is its speed when it is 0.200 m from \(q_1\) ?

In a certain region of space, the electric potential is \(V(x, y, z) = Axy - Bx^2 + Cy,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x\)-, \(y\)-, and \(z\)-components of the electric field. (b) At which points is the electric field equal to zero?

Two point charges \(q_1 = +\)2.40 nC and \(q_2 = -\)6.50 nC are 0.100 m apart. Point \(A\) is midway between them; point \(B\) is 0.080 m from \(q_1\) and 0.060 m from \(q_2\) (\(\textbf{Fig. E23.19}\)). Take the electric potential to be zero at infinity. Find (a) the potential at point \(A\); (b) the potential at point \(B\); (c) the work done by the electric field on a charge of 2.50 nC that travels from point \(B\) to point \(A\).

A metal sphere with radius \(R_1\) has a charge \(Q_1\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_2\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.
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