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Chapter 19: Problem 27

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of \(\operatorname{mass} 7.20 \times 10^{-3} \mathrm{kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

Short Answer

Expert verified
The particle travels 0.0629 m before its speed is zero.

Step by step solution

01

Understand the Concept

The problem involves two charged particles where one is fixed, and the other is moving toward it. The force involved here is electrostatic repulsion due to like charges. As the moving charge approaches the fixed charge, it will slow down until the kinetic energy converts completely into electrostatic potential energy.
02

Write Down Given Values

Given: \( q_1 = -3.00 \mu \mathrm{C} = -3.00 \times 10^{-6} \mathrm{C} \), \( q_2 = -8.00 \mu \mathrm{C} = -8.00 \times 10^{-6} \mathrm{C} \), initial distance \( r_1 = 0.0450 \mathrm{m} \), mass \( m = 7.20 \times 10^{-3} \mathrm{kg} \), and initial speed \( v = 65.0 \mathrm{m/s} \).
03

Calculate Initial Kinetic Energy

The initial kinetic energy (\( KE_{initial} \)) of the moving charge is given by \( KE_{initial} = \frac{1}{2}mv^2 \).\[ KE_{initial} = \frac{1}{2}(7.20 \times 10^{-3} \mathrm{kg})(65.0 \mathrm{m/s})^2 \approx 15.21 \mathrm{J} \]
04

Calculate Initial Potential Energy

Calculate the initial potential energy (\( PE_{initial} \)) between the two charges using the formula \( PE = \frac{k|q_1q_2|}{r_1} \). The electrostatic constant \( k = 8.99 \times 10^9 \: N \cdot m^2/C^2 \).\[ PE_{initial} = \frac{8.99 \times 10^9 \times 3.00 \times 10^{-6} \times 8.00 \times 10^{-6}}{0.0450} \approx 4.79 \mathrm{J} \]
05

Setup Energy Conservation Equation

The total mechanical energy is conserved. So, initial kinetic plus potential energy equals final potential energy when the speed is zero. \[ KE_{initial} + PE_{initial} = PE_{final} \]
06

Solve for Final Distance

Substitute the values to find the final distance \( r_2 \).\[ 15.21 \mathrm{J} + 4.79 \mathrm{J} = \frac{8.99 \times 10^9 \times 3.00 \times 10^{-6} \times 8.00 \times 10^{-6}}{r_2} \]\[ 20.00 = \frac{8.99 \times 10^9 \times 2.40 \times 10^{-11}}{r_2} \]\[ r_2 = \frac{8.99 \times 10^9 \times 2.40 \times 10^{-11}}{20.00} \approx 0.1079 \mathrm{m} \]
07

Calculate the Travel Distance

The distance the particle travels before its speed is zero is the difference between initial and final positions: \[ \Delta r = r_2 - r_1 = 0.1079 \mathrm{m} - 0.0450 \mathrm{m} = 0.0629 \mathrm{m} \]

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Key Concepts

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

Conservation of Energy
The conservation of energy principle is a fundamental concept in physics where energy in a closed system remains constant over time. It states that energy can neither be created nor destroyed; instead, it can only be transformed from one form to another. In this exercise, the system involves two charged particles where energy is transferred between kinetic energy (energy due to motion) and potential energy (energy due to position in a force field).

  • Initially, the moving particle has kinetic energy because of its velocity and potential energy due to its position relative to the fixed charge.
  • As it moves closer to the fixed charge, kinetic energy decreases while potential energy increases, conserving the total mechanical energy of the system.
  • Eventually, the particle's velocity becomes zero, meaning all of its initial kinetic energy has transformed into electrostatic potential energy.
Understanding this transfer between different energy types is essential, and applying it correctly allows us to solve when and where certain conditions, like zero velocity, are met.
Kinetic Energy
Kinetic energy is the energy that a particle possesses due to its motion. It is dependent on both the mass and speed of the particle, which is given by the formula, \( KE = \frac{1}{2}mv^2 \), where:
  • \( m \) is the mass of the particle and
  • \( v \) is its velocity.
In our problem, we start with a moving particle that has a certain mass (7.20 x 10^-3 kg) and is moving at an initial speed of 65.0 m/s. This gives us a significant amount of kinetic energy, calculated to be approximately 15.21 J.

  • This kinetic energy represents the particle's ability to overcome forces acting upon it, such as the electrostatic repulsion from the fixed charge.
  • As the distance between the particles changes, the kinetic energy will convert into potential energy.
  • When the particle reaches a point of rest, all the kinetic energy has been fully transferred to potential energy.
Kinetic energy plays a pivotal role in determining how far the particle will travel before stopping. By using its initial kinetic energy, we can predict how energy shifts occur within the problem.
Potential Energy
Potential energy in this context is the energy stored in the system due to the electrostatic forces between the charged particles. The potential energy is calculated using the formula, \( PE = \frac{k|q_1q_2|}{r} \), where:
  • \( k \) is Coulomb's constant (8.99 x 10^9 N路m虏/C虏),
  • \( q_1 \) and \( q_2 \) are the charges of the particles, and
  • \( r \) is the distance between the charges.
Initially, the particle is positioned with a measured potential energy due to the spacing between the negative charges, which repulse each other. This initial potential energy was found to be 4.79 J.

  • As the charge moves closer to the fixed charge, the potential energy increases because the electrostatic force grows stronger with decreasing distance.
  • At the moment the particle stops, its entire kinetic energy has transformed into potential energy, providing us a way to find this final resting distance through energy conservation equations.
  • The change in potential energy helps understanding how forces interact over distances and bring objects to a state of balance or zero motion.
In electrostatics, potential energy impacts how particles interact, showing potential pathways and positions of equilibrium.

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Most popular questions from this chapter

At a distance of \(1.60 \mathrm{m}\) from a point charge of \(+2.00 \mu \mathrm{C}\), there is an equipotential surface. At greater distances there are additional equipotential surfaces. The potential difference between any two successive surfaces is \(1.00 \times 10^{3} \mathrm{V} .\) Starting at a distance of \(1.60 \mathrm{m}\) and moving radially outward, how many of the additional equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not include the starting surface.

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about \(0.070 \mathrm{V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{m} .\) What is the magnitude of the electric field in the membrane?

Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.070 V. How much work is done by the electric force when a sodium ion (charge \(=+e\) ) moves from the outside to the inside?

A particle with a charge of \(-1.5 \mu \mathrm{C}\) and a mass of \(2.5 \times 10^{-6} \mathrm{kg}\) is released from rest at point \(A\) and accelerates toward point \(B\), arriving there with a speed of \(42 \mathrm{m} / \mathrm{s}\). The only force acting on the particle is the electric force. (a) Which point is at the higher potential? Give your reasoning. (b) What is the potential difference \(V_{\mathrm{B}}-V_{\mathrm{A}}\) between \(\mathrm{A}\) and \(\mathrm{B} ?\)

Particle 1 has a mass of \(m_{1}=3.6 \times 10^{-6} \mathrm{kg},\) while particle 2 has a mass of \(m_{2}=6.2 \times 10^{-6} \mathrm{kg} .\) Each has the same electric charge. These particles are initially held at rest, and the two- particle system has an initial electric potential energy of 0.150 J. Suddenly, the particles are released and fly apart because of the repulsive electric force that acts (a) Two particles have different masses, but the same electrical charge \(q\) They are initially at rest. (b) At the instant following the release of the particles, they are flying apart due to the mutual force of electric repulsion. on each one (see the figure). The effects of the gravitational force are negligible, and no other forces act on the particles. Concepts: (i) What types of energy does the two-particle system have initially? (ii) What types of energy does the two-particle system have at the instant illustrated in part \(b\) of the drawing? (iii) Does the principle of conservation of energy apply to this problem? Explain. (iv) Does the conservation of linear momentum apply to the two particles as they fly apart? Explain. Calculations: At one instant following the release, the speed of particle 1 is measured to be \(v_{1}=170 \mathrm{m} / \mathrm{s} .\) What is the electric potential energy at this instant?
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