/*! 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} Q4P In Fig. 23-32, a butterfly net i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Q4P (page 679)

In Fig. 23-32, a butterfly net is in a uniform electric field of magnitude E=3.0mN/C. The rim, a circle of radiusa=11cm, is aligned perpendicular to the field. The net contains no net charge. Find the electric flux through the netting.

Short Answer

Expert verified

The electric flux through the netting is -1.1脳10-4N.m2/C.

Step by step solution

01

The given data

  1. The magnitude of the electric field,E=3.0m.N/C
  2. Radius of the circle,a=11cm1m100cm=0.11m
  3. The net contains no charge.
02

Understanding the concept of electric flux

Using the concept of Gauss flux theorem through an enclosed surface, we can get the flux value of the electric field through a circular surface.

Formula:

The electric flux through a circular surface, 蠒=蟺a2E (1)

03

Calculation of the electric flux

The flux through the flat surface encircled by the rim is given using equation (1) and thus, the flux through the netting is given as:

蠒'=-蠒=-蟺(0.11m)2(3.0脳10-3N/C)=-1.1脳10-4N.m2/C

Hence, the value of the flux is -1.1脳10-4N.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

Three infinite non-conducting sheets, with uniform positive surface charge densities蟽,2蟽,and 3蟽,are arranged to be parallel like the two sheets in Fig. 23-19a. What is their order, from left to right, if the electric field produced by the arrangement has magnitudeE=0in one region andE=2蟽/蔚0in another region?

Figure 23-50 shows a very large non-conducting sheet that has a uniform surface charge density of 蟽=-2.00渭C/m2, it also shows a particle of chargeQ=6.00渭C, at distance dfrom the sheet. Both are fixed in place. If d=0.200m , at what (a) positive and (b) negative coordinate on the xaxis (other than infinity) is the net electric field Enet of the sheet and particle zero? (c) If d=0.800m , at what coordinate on the x-axis isEnet=0?

The net electric flux through each face of a die (singular of dice) has a magnitude in units of 103Nm2/Cthat is exactly equal to the number of spots Non the face (1 through 6). The flux is inward for Nodd and outward for Neven. What is the net charge inside the die?

A Gaussian surface in the form of a hemisphere of radiusR=5.68cmlies in a uniform electric field of magnitudeE=2.50鈥�N/C. The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. What is the flux through

(a) the base and

(b) the curved portion of the surface?

Two long, charged, thin-walled, concentric cylindrical shells have radii of3.0 cm and 6.0 cm . The charge per unit length is 5.0脳10-6C/mon the inner shell and -7.0脳10-6C/mon the outer shell. What are the (a) magnitude Eand (b) direction (radially inward or outward) of the electric field at radial distance r=4.0 cm ? What are (c) Eand (d) the direction at r=8.0 cm?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Medical Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Dynamics

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.