/*! 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 13 A particle of charge \(q\) is fi... [FREE SOLUTION] | 91影视

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Chapter 24: Problem 13

A particle of charge \(q\) is fixed at point \(P\), and a second particle of mass \(m\) and the same charge \(q\) is initially held a distance \(r_{1}\) from \(P\). The second particle is then released. Determine its momentum magnitude when it is a distance \(r_{2}\) from \(P\). Let \(q=3.1 \mu \mathrm{C}\), \(m=20 \mathrm{mg}, r_{1}=0.90 \mathrm{~mm}\), and \(r_{2}=1.5 \mathrm{~mm}\).

Short Answer

Expert verified
The momentum at distance \( r_2 \) is calculated using energy conservation principles.

Step by step solution

01

Analyze Electrostatic Potential Energy

The initial potential energy (PE) of the system when the second charged particle is at distance \( r_1 \) is given by:\[ \text{PE}_1 = \frac{k q^2}{r_1} \]The potential energy when the particle is at distance \( r_2 \) is:\[ \text{PE}_2 = \frac{k q^2}{r_2} \] Here, \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \mathrm{~N~m^2/C^2} \). We need to find the difference in potential energy to determine the kinetic energy gained.
02

Apply Conservation of Energy

By conservation of energy, the change in potential energy is equal to the change in kinetic energy:\[ \text{KE} = \text{PE}_1 - \text{PE}_2 = \frac{k q^2}{r_1} - \frac{k q^2}{r_2} \]This change in energy equals the kinetic energy of the particle at distance \( r_2 \).
03

Calculate Kinetic Energy

Substitute \( q = 3.1 \times 10^{-6} \mathrm{~C} \), \( r_1 = 0.90 \times 10^{-3} \mathrm{~m} \), \( r_2 = 1.5 \times 10^{-3} \mathrm{~m} \) into the formula:\[ \text{KE} = \frac{(8.99 \times 10^9)(3.1 \times 10^{-6})^2}{0.90 \times 10^{-3}} - \frac{(8.99 \times 10^9)(3.1 \times 10^{-6})^2}{1.5 \times 10^{-3}} \]Calculate the kinetic energy.
04

Relate Kinetic Energy to Momentum

The kinetic energy \( \text{KE} \) is related to the momentum \( p \) of the particle by the formula:\[ \text{KE} = \frac{p^2}{2m} \]Rearranging this gives:\[ p = \sqrt{2m \times \text{KE}} \]
05

Calculate the Momentum

Use \( m = 20 \times 10^{-6} \mathrm{~kg} \) and the calculated \( \text{KE} \) from Step 3 to find \( p \):\[ p = \sqrt{2 imes 20 \times 10^{-6} \times \text{KE}} \]Calculate the value to find the particle's momentum when it is distance \( r_2 \) from \( P \).

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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, stating that energy cannot be created or destroyed in an isolated system. It can only be transformed from one form to another. In the context of electrostatics, this means that as a particle moves due to electric forces, its total energy is conserved.
In our scenario, the particle starts with some potential energy because of its position relative to another charged particle. As it moves from an initial distance (\(r_1\)) to a new distance (\(r_2\)), its potential energy changes. This change results in a corresponding change in kinetic energy. It's like exchanging energy stored in position for energy of motion.
  • The initial potential energy (\( ext{PE}_1 \)) is higher because the particles are closer together, resulting in stronger electrostatic repulsion.
  • At a further distance (\(r_2\)), the potential energy (\( ext{PE}_2 \)) is lower.
  • The difference (\( ext{PE}_1 - ext{PE}_2 \)) becomes the kinetic energy (\( ext{KE} \)) gained by the particle.
The beauty of this concept is how it simplifies complex systems by allowing us to track energy transformation without worrying about the nitty-gritty details of what happens in between.
Kinetic Energy
Kinetic energy describes the energy a particle has due to its motion. It's an important part of understanding how objects behave when they are moving under the influence of forces. The kinetic energy of a particle is calculated using the formula:\(\text{KE} = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity.
In our exercise, the static interaction between two charges brings about a change in kinetic energy as the particle moves away due to electrostatic force. The potential energy difference calculated earlier becomes the kinetic energy at \(r_2\).
  • The kinetic energy reflects how much motion energy the particle has gained after being released.
  • We can infer the speed of particles from kinetic energy, highlighting how particle velocity is tied to its gained or lost energy due to forces acting upon it.
These principles help us compute the particle's momentum, combining both its speed and mass properties. Understanding kinetic energy is essential for grasping momentum changes in moving particles.
Momentum Calculation
Momentum is a measure of the motion of a particle. It is calculated by multiplying the mass of the particle by its velocity, denoted by \( p = mv \). For the particle in our problem, momentum represents both the motion quantity and its direction. The primary relationship between momentum and energy in this instance comes through kinetic energy.
  • The relationship \( ext{KE} = \frac{p^2}{2m} \) links kinetic energy and momentum.
  • By rearranging, we arrive at \( p = \sqrt{2m \times \text{KE}} \), allowing calculation of momentum once kinetic energy is known.
Thus, with potential energy converted to kinetic energy, we can determine how this energetic transition moves to the particle's momentum when it reaches the new position at \( r_2 \). By understanding this relationship, students can see how energy and motion interact, offering deeper insight into the fundamental laws governing particle dynamics.

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Most popular questions from this chapter

Identical \(50 \mu \mathrm{C}\) charges are fixed on an \(x\) axis at \(x=\pm 2.0 \mathrm{~m} .\) A particle of charge \(q=-15 \mu \mathrm{C}\) is then released from rest at a point on the positive part of the \(y\) axis. Due to the symmetry of the situation, the particle moves along the \(y\) axis and has kinetic energy \(1.2 \mathrm{~J}\) as it passes through the point \(x=0, y=4.0 \mathrm{~m}\). (a) What is the kinetic energy of the particle as it passes through the origin? (b) At what negative value of \(y\) will the particle momentarily stop?

An electron is projected with an initial speed of \(1.6 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to twice the initial value?

(a) What is the electric potential energy of two electrons separated by \(3.00 \mathrm{~nm}\) ? (b) If the separation increases, does the potential energy increase or decrease?

The electric potential \(V\) in the space between two flat parallel plates 1 and 2 is given (in volts) by \(V=1500 x^{2}\), where \(x\) (in meters) is the perpendicular distance from plate 1 . At \(x=1.8 \mathrm{~cm}\), (a) what is the magnitude of the electric field and (b) is the field directed toward or away from plate \(1 ?\)

An infinite nonconducting sheet with a uniform surface charge density sets up parallel equipotential surfaces. Any pair of surfaces differing by \(25.0 \mathrm{~V}\) are separated by \(8.80 \mathrm{~mm}\). (a) What is the magnitude of the surface charge density? (b) If an electron is released near the sheet, does it tend to move from higher to lower potential or vice versa?
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