Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 22: Problem 1

In order to use Gauss's Law to calculate the electric field created by a known distribution of charge, which of the following must be true? a) The charge distribution must be in a nonconducting medium. b) The charge distribution must be in a conducting medium. c) The charge distribution must have spherical or cylindrical symmetry. d) The charge distribution must be uniform. e) The charge distribution must have a high degree of symmetry that allows assumptions about the symmetry of its electric field to be made.

Short Answer

Expert verified
Answer: The charge distribution must have a high degree of symmetry that allows assumptions about the symmetry of its electric field to be made.

Step by step solution

01

Analyzing option (a)

The charge distribution being in a nonconducting medium is not a necessary condition for applying Gauss's Law. Gauss's Law can be used for both conducting and nonconducting media, so this option is not accurate.
02

Analyzing option (b)

Similar to option (a), Gauss's Law can be applied to both conducting and nonconducting media. Hence, this option is also not an accurate condition.
03

Analyzing option (c)

It's true that Gauss's Law becomes easier to apply in situations with charge distributions having spherical or cylindrical symmetry. However, while it makes calculations easier, it is not a strict requirement for using Gauss's Law in general. So this option is not the most accurate either.
04

Analyzing option (d)

Uniform charge distribution is not a necessary condition. Gauss's Law can be used with non-uniform charge distributions as well. As a result, this option is not accurate.
05

Analyzing option (e)

Having a high degree of symmetry in the charge distribution allows assumptions about the symmetry of its electric field to be made, which in turn simplifies the application of Gauss's Law. This is the key condition that makes Gauss's Law a powerful tool for calculating the electric field in such cases.
06

Conclusion

Based on the analysis of the given options, the most accurate statement is option (e): "The charge distribution must have a high degree of symmetry that allows assumptions about the symmetry of its electric field to be made."

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

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside its surface. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.810 \lambda\) and radius \(6.50 r ?\)

There is a uniform charge distribution of \(\lambda=6.005 \cdot 10^{-8} \mathrm{C} / \mathrm{m}\) along a thin wire of length \(L .\) The wire is then curved into a semicircle that is centered at the origin and has a radius of \(R=L / \pi .\) The magnitude of the electric field at the center of the semicircle is \(2.425 \cdot 10^{4} \mathrm{~N} / \mathrm{C}\). What is the value of \(L ?\)

An infinitely long, solid cylinder of radius \(R=9.00 \mathrm{~cm},\) with a uniform charge per unit of volume of \(\rho=6.40 \cdot 10^{-8} \mathrm{C} / \mathrm{m}^{3},\) is centered about the \(y\) -axis. Find the magnitude of the electric field at a radius \(r=4.00 \mathrm{~cm}\) from the center of this cylinder.

At which of the following locations is the electric field the strongest? a) a point \(1 \mathrm{~m}\) from a \(1-\mathrm{C}\) point charge b) a point \(1 \mathrm{~m}\) (perpendicular distance) from the center of a 1 -m-long wire with \(1 \mathrm{C}\) of charge distributed on it c) a point \(1 \mathrm{~m}\) (perpendicular distance) from the center of a \(1-\mathrm{m}^{2}\) sheet of charge with \(1 \mathrm{C}\) of charge distributed on it d) a point \(1 \mathrm{~m}\) from the surface of a charged spherical shell with a radius of \(1 \mathrm{~m}\) e) a point \(1 \mathrm{~m}\) from the surface of a charged spherical shell with a radius of \(0.5 \mathrm{~m}\) and a charge of \(1 \mathrm{C}\)

Two charges, \(+e\) and \(-e,\) are a distance of \(0.680 \mathrm{nm}\) apart in an electric field, \(E\), that has a magnitude of \(4.40 \mathrm{kN} / \mathrm{C}\) and is directed at an angle of \(45.0^{\circ}\) with respect to the dipole axis. Calculate the dipole moment and thus the torque on the dipole in the electric field.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Energy Physics

Read Explanation

Kinematics Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Magnetism and Electromagnetic Induction

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