/*! 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} Q2Q The graph in Fig.16.56 shows the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The graph in Fig.16.56 shows the electric potential energy for a system of two interacting objects, as a function of the distance between the objects. What system might this graph represent?

(1) Two Protons, (2) Two sodium ions, (3) Two neutrons, (4) Two chloride ions, (5) Two electrons, (6) A Proton and an Electron, (7) A sodium ion and a chloride ion.

Short Answer

Expert verified

The above graph represents Two Protons, Two Sodium ions, Two neutrons, Two chloride ions, Two electrons

Step by step solution

01

Significance of the electric potential energy

The electric potential energy depends on the sign convention of the two charges Q1 and Q2.

02

Explanation of the graph that represents the potential energy

The equation of the magnitude of the electric potential energy is expressed as,

Ur12=kq1×q2r12

It can be inferred from the above equation that :

(1) q1and q2 must have the same sign if U>0

(2) q1 and q2 must have the opposite sign if U<0

(3) At least one of the objects is neutral if U=0

By observation of the graph of UVsr12,it can be concluded that the graph of UVsr12is positive everywhere. So, the charges will be either both positive and negative.

Thus, the given graph might represent Two Protons, Two Sodium ions, Two neutrons, Two chloride ions, two electrons.

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

Question: An electron passes through a region in which there is an electric field, and whiles it is in the region its kinetic energy decreases by 4.5×10-17J. Initially the kinetic energy of the electron was4.5×10-17J . What is the final speed of the electron? (You can use the approximate (nonrelativistic) equation here.)

The long rod shown in Figure 16.76 has length L and carries a uniform charge −Q. Calculate the potential difference VA-VC. All of the distances are small compared to L. Explain your work carefully

Four voltmeters are connected to a circuit as shown in figure 16.90. As is usual with voltmeters, the reading on the voltmeter is positive if the negative lead (black wire, usually labled COM) is connected to a location at lower potential, and the positive lead(red) is connected to a location at higher potential. The circuit contains two devices whose identity is unknown and a rod (green) of length 9 cm made of conducting material. At a particular moment, the reading observed in the voltmeters are, voltmeter A: -1.6 V, voltmeter B: -6 V, voltmeter A: -3.5 V. (a) At this moment, what is the reading on voltmeter D, both magnitude and sign? (b) What are the magnitude and direction of the electric field inside the rod?

Two very large disks of radius Rare carrying uniformly distributed charges QAand QB. The plates are parallel and 0.1mmapart, as shown in Figure 16.70. The potential difference between the plates isVB-VA=-10V. (a) What is the direction of the electric field between the disks? (b) Invent values of QA, QBand Rthat would make VB-VA=-10V.

A capacitor with a gap of 1mm has a potential difference from one plate to the other of36V . What is the magnitude of the electric field between the plates?

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.