/*! 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} Q9DQ If you carry out the integral of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If you carry out the integral of the electric field S for a closedpath like that shown in Fig. Q23.9, the integral will alwaysbe equal to zero, independent of the shape of the path and independent of where charges may be located relative to the path. Explain why

Short Answer

Expert verified

The finite integration over a closed path with the initial value and the final value is equal, then the value of the integration is equal to zero.

Step by step solution

01

About electric field 

electric field, an electric property associated with each point in space when charge is present in any form.The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

02

Determine why the integral will always be equal to zero or independent of shape of path

As we know the electric field is a conservative filed- Also, We know that the electric field is given by the following relation:

So, we solve for the given integral on the closed path:

As we know the closed integral on the gradient of the electric potential is equal to zero. Hence, the potential initial potential

is equal to the ?nal potentiaL We know that the finite integration over a closed path with the initial value and the final value is

equal, then the value of the integration is equal to zero. 80, the work done is independent on the path and is zero for the

closed paths-

Result

As We know the closed integral on the gradient of the electric potential is equal to zero- Hence, the potential initial potential is equal to the final potential- We know that the finite integration over a closed path with the initial value and the final value is equal, then the value of the integration is equal to zero.

So, the Work done is independent on the path and is zero for the closed paths.

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 beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction pependicurlar to its original direction (Fig. E27.24). The beam travels a distance of 1.10 cm while in the field. What is the magnitude of the magnetic field?

A horizontal rectangular surface has dimensions 2.80cmby 3.20cmand is in a uniform magnetic field that is directed at an angle of 30.0°above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10×10-4Wb through the surface?

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer’s heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

Question: A conducting sphere is placed between two charged parallel plates such as those shown in Figure. Does the electric field inside the sphere depend on precisely where between the plates the sphere is placed? What about the electric potential inside the sphere? Do the answers to these questions depend on whether or not there is a net charge on the sphere? Explain your reasoning.

Each of the lettered points at the corners of the cube in Fig. Q27.12 represents a positive charge qmoving with a velocity of magnitude vin the direction indicated. The region in the figure is in a uniform magnetic field , parallel to the x-axis and directed toward the right. Which charges experience a force due to B⇶Ä? What is the direction of the force on each charge?

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.