Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 11

Why can't you use Gauss's law to determine the field of a uniformly charged cube? Why couldn't you use a cubical Gaussian surface?

Short Answer

Expert verified
Gauss's law cannot be used to determine the electric field of a uniformly charged cube because the lack of symmetry in a cube, unlike spherically or cylindrically symmetric distribution, preventing the simplification of the integral in Gauss's law. Even if a cubical Gaussian surface is used, it is impossible to have the electric field be both constant in magnitude and perpendicular to the surface at all points, a requirement for Gauss's law application.

Step by step solution

01

Understanding Gauss's Law

Gauss's law states that the net electric flux through a closed surface (Gaussian surface) is equal to \(\frac{1}{\epsilon_0}\) times the charge enclosed by the surface. This law is utilized best when there is symmetry in the charge distribution.
02

Applying Gauss's Law to a Cubically Charged Object

With a cube, the electric field vectors' direction changes significantly at different points due to the orientation of the edges, faces, and vertices. This prevents simplification of Gauss's law integral.
03

Considering the Use of a Cubical Gaussian Surface

Even if a cubical Gaussian surface is used, the electric field does not maintain equal magnitude or direction at all points on the surface due to cube's geometry as the field is not perpendicular or constant at all points. This leads to a non-zero integral, making the law unapplicable.

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 point charge \(-q\) is at the center of a spherical shell carrying charge \(+2 q .\) That shell, in turn, is concentric with a larger shell carrying \(-\frac{3}{2} q .\) Draw a cross section of this structure, and sketch the electric field lines using the convention that eight lines correspond to a charge of magnitude \(q\).

A point charge is located a fixed distance outside of a uniformly charged sphere. If the sphere shrinks in size without losing any charge, what happens to the force on the point charge?

A \(2.6-\mu \mathrm{C}\) charge is at the center of a cube \(7.5 \mathrm{cm}\) on each side. What's the electric flux through one face of the cube? (Hint: Think about symmetry, and don't do an integral.)

You're sitting inside an uncharged, hollow spherical shell. Suddenly someone dumps a billion coulombs of charge on the shell, distributed uniformly. What happens to the electric field at your location?

The electric field of a flat sheet of charge is \(\sigma / 2 \epsilon_{0} .\) Yet the field of a flat conducting sheet-even a thin one, like a piece of aluminum foil- is \(\sigma / \epsilon_{0} .\) Explain this apparent discrepancy.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Physical Quantities And Units

Read Explanation

Solid State Physics

Read Explanation

Quantum Physics

Read Explanation

Fluids

Read Explanation

Scientific Method 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