91Ó°ÊÓ

Two protons are held at rest in vacuum, \(5.0 \times 10^{-12} \mathrm{~m}\) apart. When released, they fly apart. How fast will each be moving when they are far from each other? Their original \(\mathrm{PE}_{E}\) will be changed to KE. Proceed as in Problem 25.13. The potential at \(5.0 \times 10^{-12} \mathrm{~m}\) from the first charge due to that charge alone is $$ V=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{1.60 \times 10^{-19} \mathrm{C}}{5 \times 10^{-12} \mathrm{~m}}\right)=288 \mathrm{~V} $$ The work needed to bring in the second proton is then $$ W=q V=\left(1.60 \times 10^{-19} \mathrm{C}\right)(288 \mathrm{~V})=4.61 \times 10^{-17} \mathrm{~J} $$ and this is the \(\mathrm{PE}_{E}\) of the original system. From the conservation of energy, $$ \begin{array}{l} \text { Original } \mathrm{PE}_{E}=\text { final } \mathrm{KE} \\ 4.61 \times 10^{-17} \mathrm{~J}=\frac{1}{2} m_{1} v_{1}^{2}=\frac{1}{2} m_{2} v_{2}^{2} \end{array} $$ Since the particles are identical, \(v_{1}=v_{2}=v\). Solving, we find that \(v=1.7 \times 10^{5} \mathrm{~m} / \mathrm{s}\) when the particles are far apart.

Step by step solution

01

Calculate the Potential Energy

First, we need to calculate the initial potential energy (PE) of the system due to the electrostatic interaction between the two protons. The potential energy is given by the equation: \[ PE_E = q \cdot V = (1.60 \times 10^{-19} \text{ C})(288 \text{ V}) = 4.61 \times 10^{-17} \text{ J} \] where \(q\) is the charge of a proton and \(V\) is the potential at the location of the second proton.

02

Apply Conservation of Energy

According to the conservation of energy, the initial potential energy of the system becomes the kinetic energy (KE) of the two protons when they are far apart. Thus, \[ \text{Original } PE_{E} = \text{final } KE \]\[ 4.61 \times 10^{-17} \text{ J} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \] Since the protons are identical, \( v_1 = v_2 = v \) and the equation simplifies to \[ 4.61 \times 10^{-17} \text{ J} = m v^2 \] where \( m \) is the mass of each proton.

03

Solve for Velocity

The mass of a proton is \( m = 1.67 \times 10^{-27} \text{ kg} \). Substituting this value into the equation derived from the conservation of energy gives:\[ 4.61 \times 10^{-17} \text{ J} = (1.67 \times 10^{-27} \text{ kg}) v^2 \]Solving for \( v \):\[ v^2 = \frac{4.61 \times 10^{-17} \text{ J}}{1.67 \times 10^{-27} \text{ kg}} \]\[ v = \sqrt{\frac{4.61 \times 10^{-17}}{1.67 \times 10^{-27}}} \]\[ v = 1.7 \times 10^5 \text{ m/s} \] Thus, the speed of each proton is \( 1.7 \times 10^5 \text{ m/s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Potential Energy

Potential energy (PE) is the energy a system has due to the positioning of its components. In the case of the two protons, this energy arises from their electrostatic interaction.
The potential energy between two charges can be calculated using the formula: \(PE_E = q \cdot V\), where:

• \(q\) is the charge of the proton, \(1.60 \times 10^{-19} \text{ C}\).
• \(V\) is the electric potential due to a point charge, determined by the formula \(V = \frac{k \cdot q}{r}\), where \(k\) is Coulomb's constant \(9.0 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\), and \(r\) is the distance between the charges, \(5.0 \times 10^{-12} \text{ m}\).

This potential energy represents the stored energy as a result of the protons' positions in the electric field.

Kinetic Energy

Kinetic energy (KE) is the energy associated with the motion of an object, in this case, the protons as they move apart. As the potential energy (PE) of the system decreases, the kinetic energy of the moving protons increases correspondingly.
The relationship between these energies is expressed by the formula for kinetic energy: \(KE = \frac{1}{2} m v^2\), where:

• \(m\) is the mass of a proton, \(1.67 \times 10^{-27} \text{ kg}\).
• \(v\) is the velocity of the proton.

Once the protons are released, their potential energy starts to convert into kinetic energy, which causes them to speed up as they move away from each other. This transformation occurs because energy is conserved in an isolated system.

Conservation of Energy

The conservation of energy principle states that energy within an isolated system cannot be created or destroyed, only transformed from one form to another. In the context of our exercise, this principle explains how the potential energy between the two protons is transformed into kinetic energy as they fly apart.
Initially, when the protons are held at a distance, all energy in the system is potential energy. As they are released, this potential energy converts into kinetic energy.
The conservation of energy can be mathematically expressed as: \[\text{Initial } PE_E = \text{Final } KE\]This equation ensures that the total energy remains constant throughout the process, providing a predictable and quantitative description of how proton velocities increase as they move apart.

Electrostatic Interaction

Electrostatic interaction is the interaction between electrically charged objects. It plays a central role in determining potential energy when two protons are held at a given distance.
This interaction is governed by Coulomb's law, which describes the electrostatic force between charged particles. The force is directly proportional to the product of their charges and inversely proportional to the square of the distance separating them.
In our scenario:

• The force and resultant potential energy arise from the protons having like charges, causing repulsion.
• This repelling force increases their distance, transforming potential energy into kinetic energy.

Understanding electrostatic interactions is crucial as they dictate how charged particles like protons behave in electric fields, influencing both their motion and energy transformation.