/*! 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} Q. 27 FIGURE EX24.27 shows a hollow ca... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Q. 27 (page 684)

FIGURE EX24.27 shows a hollow cavity within a neutral conductor. A point charge Qis inside the cavity. What is the net electric flux through the closed surface that surrounds the conductor?

Short Answer

Expert verified

The net electric flux through the closed surface is桅e=Q蔚0

Step by step solution

01

Given information and Theory used 

Given figure :

Theory used :

The amount of electric field that travels through a closed surface is referred to as the electric flux. The electric flux through a surface is proportional to the charge inside the surface, according to Gauss's law, which is provided by equation :

桅e=鈭�E鈫�路dA鈫�=Qin蔚0 (1)

02

Calculating the net electric flux through the closed surface 

The cavity's positive charge Qinduces a negative charge-Qon the conductor's inner side.

As seen in the diagram below, the induced negative charge produces a positive charge Qat the conductor's surface.

Qin=Q+(-Q)+Q=Qis the total enclosed charge through the closed surface that surrounds the conductor.

To calculate the electric flux, enter this number into equation (1).

桅e=Qin蔚0=Q蔚0

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 electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/系0, with Qin/系, where 系 is the permittivity of the material. (Technically,系0 is called the vacuum permittivity.) Suppose that a 50nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is2500N/C . What is the permittivity of rubber?

All examples of Gauss's law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we've claimed that the net 桅e=Qin/系0is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length Lcentered on a long thin wire with linear charge density 位. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral.

a. Consider the face parallel to the yz-plane. Define area dA鈫�as a strip of width dyand height Lwith the vector pointing in the x-direction. One such strip is located at position localid="1648838849592" y. Use the known electric field of a wire to calculate the electric flux localid="1648838912266" d桅through this little area. Your expression should be written in terms of y, which is a variable, and various constants. It should not explicitly contain any angles.

b. Now integrate d桅to find the total flux through this face.

c. Finally, show that the net flux through the cube is 桅e=Qin/系0.

A long cylinder with radius band volume charge density 蚁has a spherical hole with radius a<b centered on the axis of the cylinder. What is the electric field strength inside the hole at radial distance r<a in a plane that is perpendicular to the cylinder through the center of the hole?

The net electric flux through an octahedron is 鈭�1000Nmm2/C. How much charge is enclosed within the octahedron?

FIGURE P24.48shows two very large slabs of metal that are parallel and distance lapart. The top and bottom of each slab has surface area A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal 1has total charge localid="1648838411434" Q1=Qand metal 2has total charge localid="1648838418523" Q2=2Q. Assume Qis positive. In terms of Qand localid="1648838434998" A, determine a. The electric field strengths localid="1648838424778" E1to localid="1648838441501" E5in regions 1to 5. b. The surface charge densities localid="1648838447660" 畏uto localid="1648838454086" 畏don the four surfaces ato d.
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Thermodynamics

Read Explanation

Radiation

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Medical Physics

Read Explanation

Force

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.