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Chapter 1: Problem 13

proton is traveling horizontally to the right at \(4.50 \times 10^{6}\) \(\mathrm{m} / \mathrm{s} .\) (a) Find the magnitude and direction of theweakest electric field that can bring the proton uniformly to rest over a distance of \(3.20 \mathrm{~cm}\). (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of patt (a)?

Short Answer

Expert verified
(a) 3300 N/C, to the left; (b) \(1.42 \times 10^{-8}\,\mathrm{s}\); (c) 18 N/C, to the right.

Step by step solution

01

Understanding the Problem

We need to determine the electric field that will stop a proton moving horizontally at a speed of \(4.50 \times 10^{6}\,\mathrm{m/s}\) over a distance of \(3.20\,\mathrm{cm}\). The process involves calculating acceleration using kinematic equations and relating it to the electric field using \(F = ma\).
02

Calculate Acceleration

We can find the acceleration needed to stop the proton using the equation \(v^2 = u^2 + 2as\), where \(v = 0\,\mathrm{m/s}\) is the final velocity, \(u = 4.50 \times 10^{6}\,\mathrm{m/s}\) is the initial velocity, and \(s = 3.20\,\mathrm{cm} = 0.032\,\mathrm{m}\). Solving for \(a\):\[0 = (4.50 \times 10^{6})^2 + 2a(0.032)\]\[a = -\frac{(4.50 \times 10^{6})^2}{2 \times 0.032}\]\[a = -3.16 \times 10^{14} \mathrm{m/s^2}\]
03

Find the Electric Field

The force exerted by the electric field is \(F = ma\), where \(m\) is the proton's mass. For a proton, \(m = 1.67 \times 10^{-27}\,\mathrm{kg}\). The electric field \(E\) is defined by \(F = qE\), where \(q = 1.60 \times 10^{-19}\,\mathrm{C}\) is the proton's charge. Hence, \[E = \frac{F}{q} = \frac{ma}{q}\]\[E = \frac{1.67 \times 10^{-27} \times (-3.16 \times 10^{14})}{1.60 \times 10^{-19}}\]\[E = 3.30 \times 10^{3}\,\mathrm{N/C}\]The direction of the electric field needs to be opposite to the proton's motion, which is to the left.
04

Calculate Time to Stop

The time \(t\) taken for the proton to stop can be found using \(v = u + at\). Since \(v = 0\,\mathrm{m/s}\):\[0 = 4.50 \times 10^{6} + (-3.16 \times 10^{14})t\]\[t = \frac{4.50 \times 10^{6}}{3.16 \times 10^{14}}\]\[t = 1.42 \times 10^{-8}\,\mathrm{s}\]
05

Minimum Field for an Electron

An electron's charge is the same magnitude but opposite in sign as a proton's charge, and an electron's mass is \(9.11 \times 10^{-31}\,\mathrm{kg}\). We use the same formula as for the proton:\[E = \frac{m_e a}{q_e} = \frac{9.11 \times 10^{-31} \times 3.16 \times 10^{14}}{1.60 \times 10^{-19}}\]\[E = 1.80 \times 10^{1}\,\mathrm{N/C}\]The direction of the electric field for an electron would also be opposite to its motion, which is to the right.

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Key Concepts

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

Proton Motion
In physics, understanding the motion of a proton within an electric field is crucial for solving many problems related to particle dynamics. A proton, which has a positive charge, will experience a force when placed in an electric field. This force causes the proton to accelerate, changing its velocity over time. For example, if the proton is initially moving at a high speed, the electric field can be used to either slow it down or speed it up, depending on the direction of the field relative to the proton's motion.

To stop a proton moving at a speed of \(4.50 \times 10^{6}\,\mathrm{m/s}\) within a small distance like \(3.20\,\text{cm}\), a strong electric field is needed to exert a force opposite to the direction of motion. The force can be calculated using the mass of the proton and the acceleration required to bring it to rest. Thus, understanding the motion of protons involves dynamic equations and concepts from electromagnetism.
  • Protons have a positive charge.
  • The force on a proton is determined by the electric field strength and charge.
  • The direction of force is opposite the motion to stop it.
Electron Motion
Similar to protons, electrons also interact with electric fields, but with some key differences. Electrons carry a negative charge, which affects how they respond to an electric field. If an electric field is applied that is directed to the right, an electron will experience a force directed to the left, since opposite charges attract.

To halt an electron, the same principles apply as with a proton, but the force calculated will be in the opposite direction. Electrons have much smaller mass compared to protons. This means that under the same conditions, electrons will require a significantly weaker electric field to bring them to rest over the same distance.
  • Electrons have a negative charge.
  • They require weaker electric fields to stop due to lower mass.
  • They experience force in the opposite direction than protons under like conditions.
Kinematic Equations
Kinematic equations are powerful tools used to describe the motion of particles, like protons and electrons. These equations relate quantities such as velocity, acceleration, time, and distance.

In the context of this exercise, we employ the equation \(v^2 = u^2 + 2as\) to determine the acceleration needed to bring a particle to rest. Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is acceleration, and \(s\) is the distance over which the acceleration occurs.

The approach involves calculating acceleration, which is then used to determine the electric field necessary to apply this acceleration. These equations are central to bridging the motion dynamics with electromagnetism concepts.
  • Relate velocity, acceleration, distance, and time.
  • Enable calculation of necessary parameters for motion change.
  • A vital link between motion equations and electric field influences.
Force and Acceleration
Force and acceleration are fundamental aspects of physics that describe how objects interact and move. The force acting on a charged particle can be calculated using Newton's second law, \(F = ma\), where \(m\) is the mass and \(a\) is the acceleration.

When a charged particle like a proton or an electron is within an electric field, the force it experiences is related to this field by \(F = qE\), where \(q\) is the charge and \(E\) is the electric field strength. By combining these equations, we can solve for the electric field needed to achieve a specified acceleration, ensuring the charged particle changes speed or stops as required.

This means:
  • Force is derived from the mass and desired acceleration (from kinematics).
  • Force experienced by charged particles also depends on their charge and electric field strength.
  • The direction of force is pivotal to understanding and controlling particle motion.

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Most popular questions from this chapter

Two positive point charges \(Q\) are held fixed on the \(x\)-axis at \(x=a\) and \(x=-a\). A third positive point charge \(q\), with mass \(m\), is placed on the \(x\)-axis away from the origin at a coordinate \(x\) such that \(|x| \ll a\). The charge \(q\), which is free to move along the \(x\)-axis, is then released. (a) Find the frequency of oscillation of the charge \(q\). (Himt: Review the definition of simple harmonic motion in Section 14.2. Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots\), valid for the case \(|z|<1 .\) ) (b) Suppose instead that the charge \(q\) were placed on the \(y\)-axis at a coordinate \(y\) such that \(|y|

A thin disk with a circular hole at its center, called an annulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. P1.57). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z\)-plane, with its center at the origin. For an arbitrary point on the \(x\)-axis (the axis of the annulus), find the magnitude and direction of the electric field \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus in Fig. P1.56. (c) Show that at points on the \(x\)-axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\)-axis (but cannot move off the axis). The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

A particle has charge \(-3.00 \mathrm{nC}\). (a) Find the magnitude and direction of the electric field due to this particle at a point \(0.250 \mathrm{~m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of \(12.0 \mathrm{~N} / \mathrm{C}\) ?

Particles in a Gold Ring. You have a pure (24-karat) gold ring with mass \(19.7 \mathrm{~g}\). Gold has an atomic mass of \(197 \mathrm{~g} / \mathrm{mol}\) and an atomic number of 79. (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?

Point charges \(q_{1}=-5 \mathrm{nC}\) and \(q_{2}=+5 \mathrm{nC}\) are separated by \(3 \mathrm{~mm}\), forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(37^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.5 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m}\) ?
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