/*! 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 49 Two electrons are fixed \(2.0 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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Two electrons are fixed \(2.0 \mathrm{~cm}\) apart. Another electron is shot from infinity and stops midway between the two. What is its initial speed?

Short Answer

Expert verified
The initial speed of the electron is approximately \(4.8 \times 10^6 \text{ m/s}.\)

Step by step solution

01

Understanding the Problem

We are given two fixed electrons with a distance of \(2.0 \text{ cm}\) between them. An additional electron is shot from infinity and comes to a stop midway between the other two electrons. We need to find the initial speed of this additional electron.
02

Identify Forces and Potential at Midpoint

At the midpoint, the electron feels forces and potential energy due to both fixed electrons. Since the electron is midway, the force from both fixed electrons will be equal in magnitude but opposite, so they cancel. Therefore, we only need to account for the electric potential energy due to both fixed electrons.
03

Calculate Potential Energy at Midpoint

The electric potential energy at a distance \( r \) from a point charge is given by \( V = \frac{k \, e}{r} \). For an electron midway between two fixed electrons, the total potential energy \( V_t \) is the sum of potential energies due to both electrons: \[ V_t = 2 \cdot \frac{k \, e}{0.01 \text{ m}} \]where \( e \) is the electron charge \( 1.6 \times 10^{-19} \text{ C} \) and \( k \) is Coulomb's constant \( 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
04

Convert Potential Energy to Kinetic Energy for Initial Speed

According to energy conservation, the initial kinetic energy at infinity is equal to the potential energy at the midway point, since the electron comes to rest there: \[ \frac{1}{2} m v^2 = V_t \]where \( m \) is the mass of the electron \( 9.11 \times 10^{-31} \text{ kg} \). Solving for \( v \), we have:\[ v = \sqrt{\frac{2 V_t}{m}} \]
05

Substitute and Solve

Substitute \( V_t \) from the potential energy calculation and the mass \( m \):\[ v = \sqrt{\frac{2 \cdot \left(2 \cdot \frac{9 \times 10^9 \cdot 1.6 \times 10^{-19}}{0.01}\right)}{9.11 \times 10^{-31}}} \]Calculate this to find \( v \), the initial speed of the electron.

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Key Concepts

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

Coulomb's Law and Distance Dependence
Coulomb's law is fundamental in understanding the forces between charged particles. It states that the electric force
  • between two point charges is directly proportional to the product of their charges
  • and inversely proportional to the square of the distance between them.
The formula for Coulomb's law is:\[ F = \frac{k \times |q_1 \times q_2|}{r^2} \]Here, \( k \) is Coulomb’s constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
This law helps in calculating the forces that charged particles exert on each other.When two charges are equal and opposite, like the electrons in this problem, the forces do not always result in motion as they cancel each other out. Instead, they contribute to the electric potential energy. The distance between charges plays a crucial role since the force decreases as distance increases.In the exercise, even though the forces cancel out at the midpoint, they add up to affect the electron shot from infinity by modifying its potential energy.
Understanding Kinetic Energy Conversion
The concept of kinetic energy is central to solving this problem. Kinetic energy is the energy of motion, defined by the formula:\[ KE = \frac{1}{2} m v^2 \]where \( m \) is mass and \( v \) is velocity. When the electron is shot from infinity, it initially possesses kinetic energy that decreases as it converts into electric potential energy, explained by energy conservation.
  • Initially, all its energy is kinetic when it's far away from the other electrons.
  • As it travels closer, it loses kinetic energy since this energy is converted into potential energy.
  • Finally, it stops at the midpoint, where all its energy is electric potential energy.
By equating the initial kinetic energy to the potential energy at the stopping point, we solve for its initial velocity. It demonstrates how energy changes from one form to another, which is a fundamental concept in physics.
Interactions Between Charges in Electric Fields
In the realm of electrostatics, charge interactions are crucial for understanding how forces and energies manifest.
  • Like charges repel each other, while unlike charges attract.
  • The interaction strength is governed by Coulomb's law, as discussed earlier.
Electric fields, created by charges, influence other charges within the field. In the exercise, the two fixed electrons produce an electric field that affects the electron shot from infinity.

The electron's movement is a result of these interactions:
  • The forces from the fixed electrons tend to cancel, but the potential energy due to their fields remains significant.
  • This potential energy ultimately dictates the electron's behavior, halting its motion at the midpoint.
Understanding these interactions helps in visualizing how electrons move through fields, a key aspect in fields like electronics and electromagnetism.

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Most popular questions from this chapter

Two uniformly charged, infinite, nonconducting planes are parallel to a \(y z\) plane and positioned at \(x=-50 \mathrm{~cm}\) and \(x=+50\) \(\mathrm{cm} .\) The charge densities on the planes are \(-50 \mathrm{nC} / \mathrm{m}^{2}\) and \(+25\) \(\mathrm{nC} / \mathrm{m}^{2}\), respectively. What is the magnitude of the potential difference between the origin and the point on the \(x\) axis at \(x=+80 \mathrm{~cm} ?\) (Hint: Use Gauss' law.)

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by \(-1.0 \mathrm{~V}\) during one revolution. Assuming the shuttle is a sphere of radius \(10 \mathrm{~m}\), estimate the amount of charge it collects.

The ammonia molecule \(\mathrm{NH}_{3}\) has a permanent electric dipole moment equal to \(1.47 \mathrm{D}\), where \(1 \mathrm{D}=1\) debye unit \(=\) \(3.34 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} .\) Calculate the electric potential due to an ammonia molecule at a point \(52.0 \mathrm{~nm}\) away along the axis of the dipole. \((\) Set \(V=0\) at infinity. \()\)

A thin, spherical, conducting shell of radius \(R\) is mounted on an isolating support and charged to a potential of \(-125 \mathrm{~V}\). An electron is then fired directly toward the center of the shell, from point \(P\) at distance \(r\) from the center of the shell \((r>R)\). What initial speed \(v_{0}\) is needed for the electron to just reach the shell before reversing direction?

A charge of \(1.50 \times 10^{-8} \mathrm{C}\) lies on an isolated metal sphere of radius \(16.0 \mathrm{~cm}\). With \(V=0\) at infinity, what is the electric potential at points on the sphere's surface?

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