/*! 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} Q6P When an electron moves from A to... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When an electron moves from A to B along an electric field line in Fig. 24-34, the electric field does 3.94 x 10-19 Jof work on it. What are the electric potential differences (a) VB - VA, (b) VC - VA, and (c) VC - VB?

Short Answer

Expert verified
  1. The electric potential difference is VB - VA = 2.46 V.
  2. The electric potential difference is VC - VA = 2.46 V.
  3. The electric potential difference is VC - VB = 0.

Step by step solution

01

Given data:

The electric field energy is 3.94 x 10-19 J.

02

Understanding the concept:

Using the equation of potential energy, you can find the electric potential difference.

∆U=e∆V

Here, ∆U is the change in potential energy, e is the electric charge having a value 1.60 x 10-19 C, and ∆Vis the electric potential difference.

03

(a) Calculate VB - VA :

The electric potential difference between A and B is define as below.

VB-VA=∆Uq=-W-e=-3.94×10-19J-1.60×10-19C=2.46V

04

(b) Calculate VC - VA:

The electric potential difference between A and C is define as below.

VC-VA=VB-VA=2.46V

05

(c) Calculate VC - VB :

Since C and B are on the same equipotential line, the potential difference between C and B is,

VC - VB = 0

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!

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

Suppose Nelectrons can be placed in either of two configurations. In configuration 1, they are all placed on the circumference of a narrow ring of radius Rand are uniformly distributed so that the distance between adjacent electrons is the same everywhere. In configuration 2, N-1electrons are uniformly

distributed on the ring and one electron is placed in the center of the ring. (a) What is the smallest value of Nfor which the second configuration is less energetic than the first? (b) For that value of N, consider any one circumference electron—call it eo. How many other circumference electrons are closer to eothan the central electron is?

Question: The ammonia molecule NH3 has a permanent electric dipole moment equal to,1.47 D where,1Debye=3.34×10-30C.m. Calculate the electric potential due to an ammonia molecule at a point 52.0nmaway along the axis of the dipole. (Set v = 0at infinity.)

A particular 12Vcar battery can send a total charge of 84 A.h(ampere-hours) through a circuit, from one terminal to the other. (a) How many coulombs of charge does this represent? (Hint:See Eq. 21-3.) (b) If this entire charge undergoes a change in electric potential of 12 V, how much energy is involved?

Particle 1 (with a charge of +5.0μC) and particle 2 (with a charge of+3.0μC) are fixed in place with separationd=4.0cmon the xaxis shown in Fig. 24-58a. Particle 3 can be moved along the xaxis to the right of particle 2. Figure 24-58bgives the electric potential energy Uof the three-particle system as a function of the xcoordinate of particle 3. The scale of the vertical axis is set byUS=5.0J.

What is the charge of particle 3?

In the situation of Question 7, is the work done by your force positive, negative, or zero if the particle is moved (a) from Ato B, (b) from Ato C, and (c) from Bto D? (d) Rank those moves according to the magnitude of the work done by your force, greatest first.

See all solutions

Recommended explanations on Physics Textbooks

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.