Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 40

An electron passes point \(A\) moving at \(6.5 \mathrm{Mm} / \mathrm{s}\). At point \(B\) it's come to a stop. Find the potential difference \(\Delta V_{A B}\)

Short Answer

Expert verified
The potential difference \(ΔV_{AB}\) can be found by assuming conservation of energy- the initial kinetic energy of the electron is converted entirely into electrical potential energy due to \(ΔV_{AB}\). Then solve for \(ΔV_{AB}\) using the equation \(ΔV_{AB} = \frac{1}{e} × \frac{1}{2} m v^2\).

Step by step solution

01

Calculation of Initial Kinetic Energy

The kinetic energy \(K.E.\) of an object moving with a speed \(v\) is given by the formula \(K.E. = \frac{1}{2} m v^2\). Here, the mass \(m\) of an electron is a known constant- it's given by \(9.11 × 10^{-31} kg\). The speed \(v\) of the electron is given as \(6.5 Mm/s\), which is \(6.5 × 10^6 m/s\). Plugging these values into the formula, we get \(K.E. = \frac{1}{2} × (9.11 × 10^{-31} kg) × (6.5 × 10^6 m/s)^2\).
02

Conversion of Kinetic Energy to Potential Energy

As the electron ends at rest, all of the kinetic energy has been transformed into electric potential energy because of the potential difference \(ΔV_{AB}\). The potential energy \(P.E.\) gained by it is equal to the work done against the electric field, or \(P.E.= e × ΔV\), where \(e\) is the charge of an electron- given by \(1.6 × 10^{-19} C\). Therefore, the initial kinetic energy of the electron is equal to its final potential energy- we have \(\frac{1}{2} m v^2 = e × ΔV_{AB}\). This gives us the equation to calculate the potential difference.
03

Calculation of Potential Difference

Rearranging the equation from step 2 for \(ΔV_{AB}\) we get \(ΔV_{AB} = \frac{1}{e} × \frac{1}{2} m v^2\). Substituting the known values into this equation will give us the required potential difference.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 disk of radius \(a\) carries nonuniform surface charge density \(\sigma=\sigma_{0}(r / a),\) where \(\sigma_{0}\) is a constant. (a) Find the potential at an arbitrary point \(x\) on the disk axis, where \(x=0\) is the disk center. (b) Use the result of (a) to find the electric field on the disk axis, and (c) show that the field reduces to an expected form for \(x \gg a\)

A sphere of radius \(R\) carries a nonuniform but spherically symmetric volume charge density that results in an electric field in the sphere given by \(\vec{E}=E_{0}(r / R)^{2} \hat{r},\) where \(E_{0}\) is a constant. Find the potential difference from the sphere's surface to its center.

"Cherry picker" trucks for working on power lines often carry electrocution hazard signs. Explain how this hazard arises and why it might be more of a danger to someone on the ground than to a worker on the truck.

Why can a bird perch on a high-voltage power line without getting electrocuted?

Can equipotential surfaces intersect? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Mechanics and Materials

Read Explanation

Fluids

Read Explanation

Force

Read Explanation

Engineering Physics

Read Explanation

Circular Motion and Gravitation

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.

Sign-up for free