/*! 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 42 As an electron moves through a r... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 42

As an electron moves through a region of space, its speed decreases from $8.50 \times 10^{6} \mathrm{m} / \mathrm{s}\( to \)2.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ The electric force is the only force acting on the electron. (a) Did the electron move to a higher potential or a lower potential? (b) Across what potential difference did the electron travel?

Short Answer

Expert verified
Answer (b): The potential difference across which the electron traveled can be found by plugging the values calculated in the previous steps into the formula for \(\Delta V\).

Step by step solution

01

Understand the work-energy principle and electron motion in an electric field

The work-energy principle states that the work done on an object by external forces equals the change in its kinetic energy. For an electron moving in an electric field, the work done by the electric force causes a change in its kinetic energy which results in a change in potential energy and electric potential.
02

Calculate the change in kinetic energy

We need to find the change in kinetic energy for the electron. The kinetic energy of an electron can be calculated by the formula \(\frac{1}{2}mv^2\), where \(m\) is the mass of the electron and \(v\) is its speed. The change in kinetic energy is then given by \(\Delta KE = KE_f - KE_i\). Given that the initial speed of the electron is \(8.5\times 10^6\) m/s and the final speed is \(2.5\times 10^6\) m/s, we calculate the change in kinetic energy using the mass of an electron, \(m = 9.11\times 10^{-31}\) kg. \( \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2\)
03

Calculate the electric force's work

The work done by the electric force on the electron is equal to the change in its kinetic energy, which is negative since the electron's speed decreases. So, \(W = \Delta KE\).
04

Determine the direction of the electric force and potential

As the electric force is the only force acting on the electron, we can determine its direction by observing the change in speed. As the electron's speed decreases, the electric force must act opposite to the motion of the electron. The electric force on an electron is given by \(F = -eE\), where \(e\) is the electron's charge and \(E\) is the electric field. Since force and electric field are in the same direction, we can conclude that the electron moves towards a higher potential (opposite to electric field direction). Answer (a): The electron moves to a higher potential.
05

Calculate the potential difference

The relation between work done by electric force and potential difference is \(W = -e\Delta V\), where \(\Delta V\) is the potential difference. As the work done is equal to the change in kinetic energy, we have: \(-e\Delta V = \Delta KE\) Now, we can rearrange for \(\Delta V\): \(\Delta V=-\frac{\Delta KE}{e}\) The electron's charge is \(e=1.6\times 10^{-19}\) C. We can calculate the potential difference using the value of \(\Delta KE\) we obtained earlier: \(\Delta V=-\frac{\Delta KE}{1.6\times 10^{-19}}\) Finally, we compute the potential difference.
06

Final Answer

(a) The electron moves to a higher potential. (b) The potential difference across which the electron traveled can be determined by plugging the values calculated in the previous steps into the formula for \(\Delta V\).

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 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
A 2.0 - \(\mu\) F capacitor is connected to a \(9.0-\mathrm{V}\) battery. What is the magnitude of the charge on each plate?
Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?
A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. If the sheets of aluminum are \(0.30 \mathrm{m}\) by $0.40 \mathrm{m}$ and the waxed paper, of slightly larger dimensions, is of thickness \(0.030 \mathrm{mm}\) and dielectric constant \(\kappa=2.5,\) what is the capacitance of this capacitor?
The plates of a 15.0 - \(\mu \mathrm{F}\) capacitor have net charges of $+0.75 \mu \mathrm{C}\( and \)-0.75 \mu \mathrm{C},$ respectively. (a) What is the potential difference between the plates? (b) Which plate is at the higher potential?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Translational Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Fluids

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Physical Quantities And Units

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.