Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Q25PE (page 697)

What is the potential 0.530×10-10mfrom a proton (the average distance between the proton and electron in a hydrogen atom)?

Short Answer

Expert verified

The potential at 0.530×10-10mfrom a proton is V=27.2V.

Step by step solution

01

Given Data

Distance from proton is0.530×10-10m

02

Concept Introduction

Electric Potential Due to a Single Charge: To determine the electric potential V due to a single source charge Q at a specific location, place a test charge qat that location and determine the electric potential energy UQqof a system containing the test charge and the source charge that creates the field. The electric potential is given as–

V=UQqq=kQr...........................(1)

Where k=8.99×109N·m2/Cis a proportionality constant called Coulomb's constant.

03

Calculation for Electric Potential

The electric potential a distance r from the proton is found from Equation (1) –

V=kqr

Entering the values for k, q, and r–

V=(8.99×109N×m2/C)(1.60×10-19C)0.530×10-10m=27.2V

Therefore, the potential is obtained as V=27.2V.

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

In nuclear fission, a nucleus splits roughly in half. (a) What is the potential \(2.00 \times {10^{ - 14}}{\rm{ }}m\) from a fragment that has \(46\) protons in it? (b) What is the potential energy in \(MeV\) of a similarly charged fragment at this distance?

Sketch the equipotential lines in the vicinity of the negatively charged conductor in Figure 19.29. How will these equipotentials look a long distance from the object?

Use the characteristics of the Coulomb force to explain why capacitance should be proportional to the plate area of a capacitor. Similarly, explain why capacitance should be inversely proportional to the separation between plates.

An electron is to be accelerated in a uniform electric field having a strength of \({\bf{2}}.{\bf{00}} \times {\bf{1}}{{\bf{0}}^6}{\rm{ }}{\bf{V}}/{\bf{m}}\). (a) What energy in \(keV\) is given to the electron if it is accelerated through \({\bf{0}}.{\bf{400}}{\rm{ }}{\bf{m}}\)? (b) Over what distance would it have to be accelerated to increase its energy by \({\bf{50}}.{\bf{0}}{\rm{ }}{\bf{GeV}}\)?

The electric field strength between two parallel conducting plates separated by\(4.00\;cm\) is\(7.50 \times {10^4}\;V/m\).

(a) What is the potential difference between the plates?

(b) The plate with the lowest potential is taken to be at zero volts. What is the potential\(1.00\;cm\) from that plate (and\(3.00\;cm\) from the other)?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Translational Dynamics

Read Explanation

Quantum Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Particle Model of Matter

Read Explanation

Medical Physics

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