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A proton \(\left(q=e, m_{p}=1.67 \times 10^{-27} \mathrm{~kg}\right)\) is accelerated from rest through a potential difference of \(1.0 \mathrm{MV}\). What is its final speed?

Short Answer

Expert verified
The final speed of the proton is approximately \(1.38 \times 10^7 \text{ m/s}\).

Step by step solution

01

Understand the Given Information

We have a proton with a charge of \(q = e\) and a mass of \(m_p = 1.67 \times 10^{-27} \text{ kg}\). It is accelerated through a potential difference of \(1.0 \text{ MV}\). This means the proton gains kinetic energy from the electric potential energy.
02

Identify the Energy Conversion

The proton converts electric potential energy into kinetic energy. The work done on the charge is equal to the change in kinetic energy: \(W = qV\). Since the proton starts from rest, its initial kinetic energy is zero, so \(W = \frac{1}{2}mv^2\).
03

Use the Work-Energy Principle

The work done on the proton is given by: \(W = qV\). With \(q = e = 1.6 \times 10^{-19} \text{ C}\) and \(V = 1.0 \times 10^6 \text{ V}\), we have:\[W = (1.6 \times 10^{-19} \text{ C})(1.0 \times 10^6 \text{ V}) = 1.6 \times 10^{-13} \text{ J}\]
04

Calculate Final Kinetic Energy

The kinetic energy \(K\) of the proton at its final speed is given by \(\frac{1}{2}mv^2 = 1.6 \times 10^{-13} \text{ J}\). We'll solve for \(v\) in the next step.
05

Solve for Final Speed

Using the equation for kinetic energy, solve for speed:\[\frac{1}{2}m_pv^2 = 1.6 \times 10^{-13}\]\[v^2 = \frac{2 \times 1.6 \times 10^{-13}}{1.67 \times 10^{-27}}\]\[v = \sqrt{\frac{3.2 \times 10^{-13}}{1.67 \times 10^{-27}}}\]\[v \approx 1.38 \times 10^7 \text{ m/s}\]
06

Check Your Calculations

Verify each step for arithmetic accuracy and ensure that units are consistent. The final speed of the proton is calculated to be approximately \(1.38 \times 10^7 \text{ m/s}\).

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Key Concepts

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

Potential Difference
The potential difference, also known as voltage, is a fundamental concept in electromagnetism. It represents the difference in electric potential between two points in a circuit.
When a proton is accelerated through a potential difference, it experiences an electric force that does work on it. This work transforms potential energy into kinetic energy. The unit of potential difference is the volt (V). A potential difference of 1.0 MV (megavolt) means the proton is subjected to a voltage of one million volts, which leads to a significant increase in its kinetic energy as it accelerates through this potential difference.
Kinetic Energy
Kinetic energy is the energy that a body possesses due to its motion. It is a scalar quantity and is calculated using the equation:
\[ K = \frac{1}{2}mv^2 \]
where \( m \) is the mass and \( v \) is the speed of the object. For a proton that starts from rest and is accelerated by a potential difference, its kinetic energy increases from zero. This increase in kinetic energy is due to the work done by the electric field, converting the potential energy into the kinetic form.
Work-Energy Principle
The work-energy principle is a fundamental concept that connects work and energy. This principle states that the work done on an object is equal to the change in its kinetic energy. In the context of accelerating a proton:
  • The electric field does work on the proton as it moves through a potential difference.
  • This work is the product of the charge of the proton and the potential difference.
Mathematically, this can be expressed as:
\[ W = qV = \Delta K \]
where \( W \) is the work done, \( q \) is the charge of the proton, and \( V \) is the potential difference. This shows how the electric potential energy converts directly into kinetic energy.
Electric Potential Energy
Electric potential energy is the energy that a charged object possesses due to its position in an electric field. When a charged particle like a proton is placed in an electric field, it has potential energy due to the electric forces acting on it.
As the proton moves from a region of high potential to a region of low potential, this energy is converted into kinetic energy. The potential energy lost by the proton is directly proportional to the charge of the proton \( q \) and the potential difference \( V \). This is formulated as:
\[ ext{Potential Energy Change} = qV \]
This concept helps explain how protons are accelerated in electric fields, transferring potential energy into motion.
Charge of a Proton
The charge of a proton is a fundamental and invariant property of protons. It is a measure of the proton's ability to exert force on other charges. The charge of a proton is symbolized by \( e \) and has a positive value of approximately \( 1.6 \times 10^{-19} \text{ C} \) (Coulombs).
This charge is crucial in the context of electrostatic interactions and energy transformations.
  • In electrostatic fields, the force on the proton is directly proportional to its charge.
  • The work done on the proton, transforming its electric potential energy into kinetic energy during acceleration, is calculated using its charge.
Thus, knowing the charge of the proton allows us to understand how it interacts with electric fields and how it is accelerated by them.

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Most popular questions from this chapter

Imagine a \(+40.0-n C\) point charge in vacuum. What is the value of the electric potential \(112 \mathrm{~cm}\) away?

A charge of \(0.20 \mu \mathrm{C}\) is \(30 \mathrm{~cm}\) from a point charge of \(3.0 \mu \mathrm{C}\) in vacuum. What work is required to bring the \(0.20-\mu \mathrm{C}\) charge \(18 \mathrm{~cm}\) closer to the \(3.0-\mu \mathrm{C}\) charge?

What is the speed of a \(400 \mathrm{eV}(a)\) electron, \((b)\) proton, and \((c)\) alpha particle? In each case we know that the particle's kinetic energy is $$ \frac{1}{2} m v^{2}=(400 \mathrm{eV})\left(\frac{1.60 \times 10^{-19} \mathrm{~J}}{1.00 \mathrm{eV}}\right)=6.40 \times 10^{-17} \mathrm{~J} $$ Substituting \(m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\) for the electron, \(m_{p}=1.67 \times\) \(10^{-27} \mathrm{~kg}\) for the proton, and \(m_{\alpha}=4\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\) for the alpha particle gives their speeds as (a) \(1.186 \times 10^{7} \mathrm{~m} / \mathrm{s},\left(\right.\) b) \(2.77 \times 10^{5}\) \(\mathrm{m} / \mathrm{s}\), and \(\left(\right.\) c) \(1.38 \times 10^{5} \mathrm{~m} / \mathrm{s}\)

The nucleus of a tin atom in vacuum has a charge of \(+50 e .(a)\) Find the absolute potential \(V\) at a radial distance of \(1.0 \times 10^{-12} \mathrm{~m}\) from the nucleus. ( \(b\) ) If a proton is released from this point, how fast will it be moving when it is \(1.0 \mathrm{~m}\) from the nucleus? (a) \(V=k_{0} \frac{q}{f}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{c}^{2}\right) \frac{(50)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{10^{-12} \mathrm{~m}}=72 \mathrm{kV}\) (b) The proton is repelled by the nucleus and flies out to infinity. The absolute potential at a point is the potential difference between the point in question and infinity. Hence, there is a potential drop of \(72 \mathrm{kV}\) as the proton flies to infinity. Usually we would simply assume that \(1.0 \mathrm{~m}\) is far enough from the nucleus to consider it to be at infinity. But, as a check, compute \(V\) at \(r=1.0 \mathrm{~m}\) : $$ V_{1 \mathrm{~m}}=k_{0} \frac{q}{r}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{(50)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{1.0 \mathrm{~m}}=7.2 \times 10^{-8} \mathrm{~V} $$ which is essentially zero in comparison with \(72 \mathrm{kV}\). As the proton falls through \(72 \mathrm{kV}\), $$ \begin{aligned} \text { KE gained } &=\mathrm{PE}_{E} \text { lost } \\ \frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} &=q V \\ \frac{1}{2}\left(1.67 \times 10^{-27} \mathrm{~kg}\right) v_{f}^{2}-0 &=\left(1.6 \times 10^{-19} \mathrm{C}\right)(72000 \mathrm{~V}) \end{aligned} $$ from which \(v_{f}=3.7 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Two metal plates are attached to the two terminals of a 1.50-V battery. How much work is required to carry \(\mathrm{a}+5.0-\mu \mathrm{C}\) charge across the gap \((a)\) from the negative to the positive plate, \((b)\) from the positive to the negative plate?

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