/*! 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} Q75P The charges of an electron and a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The charges of an electron and a positron are -eand+e. The mass of each is9.11×10−31 k²µ.What is the ratio of the electrical force to the gravitational force between an electron and a positron?

Short Answer

Expert verified

The ratio of the electric force between the electron and the positron is 4.16×1042.

Step by step solution

01

Given

The mass of electron and positron is 9.11×10−31kg.

02

 Step 2: Determine the formulas for the electric forces:

Electric force between two chargesQ1andQ2is given byFe=KQ1Q2r2.

Gravitational force between two masses and is given by Fg=Gm1m2r2.

03

Calculate the ratio of the electrical force to the gravitational force between an electron and a positron?

The electrical force between an electron and a positron separated by a distance r isFe=ke2r2

On the other hand, the gravitational force between the two charges isFg=Gm2r2. Thus, the ratio of the two forces is,

FeFg=ke2r2Gm2r2=ke2Gm2

Substitute the values and solve as:

FeFg=(9×109 Nm2C2)(1.60×10−19 C)2(6.67×10−11 Nm2kg2)(9.11×10−31 k²µ)2=4.16×1042

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

The charges and coordinates of two charged particles held fixed in an x-yplane are q1=+3.0mC,x1=3.5cm,y1=0.50cm,and q2=-4.0mC,x2=-2.0cm,y2=1.5 cm.Find the (a) magnitude and (b) direction of the electrostatic force on particle 2 due to particle 1. At what (c) xand (d) ycoordinates should a third particle of charge q3=+4.0 mC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

Two equally charged particles are held3.2×10-3m apart and then released from rest. The initial acceleration of the first particle is observed to be 7.0m/s2 and that of the second to be 9.0m/s2. If the mass of the first particle is6.3×10-7kg,, what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?

In Fig. 21-24, three identical conducting spheres initially have the following charges: sphere A,4Q ; sphere B,−6Q ; and sphere C,0 . Spheres Aand Bare fixed in place, with a center-to-center separation that is much larger than the spheres. Two experiments are conducted. In experiment 1, sphere Cis touched to sphere Aand then (separately) to sphere B, and then it is removed. In experiment 2, starting with the same initial states, the procedure is reversed: Sphere Cis touched to sphere Band then (separately) to sphere A, and then it is removed. What is the ratio of the electrostatic force between Aand Bat the end of experiment 2 to that at the end of experiment 1?

In Fig. 21-38, particle 1 of charge +4eis above a floor by distance d1=2.00mmand particle 2 of charge+6eis on the floor, at distanced2=6.00mmhorizontally from particle 1.What is the xcomponent of the electrostatic force on particle 2 due to particle 1?

An electron is in a vacuum near Earth’s surface and located aty=0on a vertical yaxis. At what value of yshould a second electronbe placed such that its electrostatic force on the first electron balancesthe gravitational force on the first electron?

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.