/*! 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} Q72P The magnitude E of an electric f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The magnitude E of an electric field depends on the radial distance r according toE=A/r4, where is a constant with the unit volt–cubic meter. As a multiple of A, what is the magnitude of the electric potential difference betweenr=2.00mandr=3.00m?

Short Answer

Expert verified

The magnitude of the electric potential difference isV(r)=+2.93×10−2A .

Step by step solution

01

Given data:

The intensity of the electric fieldE=Ar4,

The value of distance are r1=2.0mand r2=3.0m

02

Understanding the concept

Use the relation between the electric field E and potential Vas given below.

role="math" localid="1662730170007" E=−dVdr

Here,r is the distance.

03

Calculate the magnitude of the electric potential difference between r= 2.00 m and r= 3.00 m:

Since, the electric field is,

E=−dVdr∫∞rdV=−∫ηnEdrV(r)−V(∞)=∫ηnAr4dr

As V∞=0, then

V(r)=−Ar3−3r1r2=−A31r23−1r13=−A3133−123=+2.93×10−2A

V(r)=+A(0.029/m3)

Hence,the magnitude of the electric potential difference is +2.93×10−2A.

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

(a) Figure 24-42ashows a non-conducting rod of length L = 6.00cmand uniform linear charge density λ=(3.68pC/m). Assume that the electric potential is defined to be V = 0at infinity. What is Vat point Pat distance d = 8.00cmalong the rod’s perpendicular bisector? (b) Figure 24-42bshows an identical rod except that one half is now negatively charged. Both halves have a linear charge density of magnitude 3.68pC/m. With V = 0at infinity, what is the net electric potential at the

VatP?

A particle of charge+7.5μ°äis released from rest at the pointx=60cmon an x-axis. The particle begins to move due to the presence of a charge Qthat remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cmif (a)Q=+20μ°äand (b)Q=-20μ°ä?

In the rectangle of Fig. 24-55, the sides have lengths 5.0 cmand15 cm, q1= -5.0 mC, and q2= +2.0 mC. With V=0at infinity, what is the electric potential at (a) corner Aand (b) corner B? (c) How much work is required to move a charge q3= +3.0 mCfrom Bto Aalong a diagonal of the rectangle? (d) Does this work increase or decrease the electric potential energy of the three-charge system? Is more, less, or the same work required if q3 is moved along a path that is (e) inside the rectangle but not on a diagonal and (f) outside the rectangle?

Question: What is the magnitude of the electric field at the point (3.00i^-2.00j^+4.00k^)m?

If the electric potential in the region is given byV=2.00xyz2, where Vis in volts and coordinates x, y, and zare in meters?

Question: What is the escape speedfor an electron initially at rest on the surface of a sphere with a radius ofand a uniformly distributed charge of1.6×10-15C? That is, what initial speed must the electron have in order to reach an infinite distance from the sphere and have zero kinetic energy when it gets there?

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.