/*! 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} Q7P A particle of charge 1.8渭颁聽is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Q7P (page 679)

A particle of charge 1.8渭颁is at the center of a Gaussian cube55cmon edge. What is the net electric flux through the surface?

Short Answer

Expert verified

The net electric flux through the surface is 2.0脳105N.m2/C.

Step by step solution

01

The given data

  1. Charge of the particle,q=1.8渭颁
  2. Edge length of cube,a=55cm1m100cm=0.55m
02

Understanding the concept of Gauss law-planar symmetry

Using the concept of Gauss law and the planar symmetry, we can get the required values of the electric field at the left of the plates, right of the plates, and between the plates.

Formula:

The total flux through any surface, 蠒=q蔚0 (1)

03

Calculation of the net flux through any cube surface

As the cube has six surfaces, thus, the net flux through each surface of the cube is given using equation (1) such that,

蠒=1.8脳10-6C6脳8.85脳10-12C2/N.m2=2.0脳105N.m2/C

Hence, the value of the required flux is2.0脳105N.m2/C

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 91影视!

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

The volume charge density of a solid nonconducting sphere of radiusR=5.60cm varies with radial distance ras given by 蚁=(14.1pC/m3)r/R. (a) What is the sphere鈥檚 total charge? What is the field magnitude E, at(b), (c) r=R/2.00, and (d) r=R? (e) Graph Eversusr.

In Fig. 23-48a, an electron is shot directly away from a uniformly charged plastic sheet, at speed V5=2.0脳105m/s . The sheet is non-conducting, flat, and very large. Figure 23-48bgives the electron鈥檚 vertical velocity component vversus time tuntil the return to the launch point. What is the sheet鈥檚 surface charge density?

Figure 23-51 shows a cross-section through a very large non-conducting slab of thicknessd=9.40mmand uniform volume charge density p=5.80fC/m3 . The origin of an x-axis is at the slab鈥檚 center. What is the magnitude of the slab鈥檚 electric field at an xcoordinate of (a) 0 , (b) 2.0mm , (c) 4.70mm , and (d) 26.0mm?

When a shower is turned on in a closed bathroom, the splashing of the water on the bare tub can fill the room鈥檚 air with negatively charged ions and produce an electric field in the air as great as 1000N/C. Consider a bathroom with dimensions2.5m脳3.0m脳2.0m. Along the ceiling, floor, and four walls, approximate the electric field in the air as being directed perpendicular to the surface and as having a uniform magnitude of600N/C. Also, treat those surfaces as forming a closed Gaussian surface around the room鈥檚 air. What are (a) the volume charge density r and (b) the number of excess elementary charges eper cubic meter in the room鈥檚 air?

In Fig. 23-43, short sections of two very long parallel lines of charge are shown, fixed in place, separated by L=8.00 cmThe uniform linear charge densities are+6.0渭颁/mfor line 1 and-2.0渭颁/mfor line 2. Where along the x-axis shown is the net electric field from the two lines zero?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Modern Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Quantum Physics

Read Explanation

Solid State 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.