/*! 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. 9 Figure EX26.9 shows a graph of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Figure EX26.9shows a graph of Vversusxin a region of space. The potential is independent of y and z. What is Exat

(a) x=-2cm, (b) x=0cm, and (c) x=2cm?

Short Answer

Expert verified

a.Ex=-2=5V/m

b.Ex=0cm=-10V/m

c.Ex=2cm=5V/m

Step by step solution

01

Given information and formula used   

Given graph :

Positions : (a) x=-2cm, (b) x=0cm, and (c) x=2cm

Theory used :

The electric field strength is assumed to be independent of both yandzpositions, i.e., it only depends on xposition. As a result,

E=-dVdx

02

Calculating Ex at (a) x = -2 cm, (b) x = 0 cm, and (c) x = 2 cm

Finding the derivative of the potential at the stated points of interest, we have :

a. At a distance of x=-2cm, the electric field strength will be :

Ex=-2cm=-ddxVx=-2cm=--10-0-1-(-3)=5V/m

b. At position x=0cm, the electric field strength will be

Ex=0cm=-ddxVx=0cm=-10-(-10)1-(-1)=-10V/m

c. At location x=2cm, the electric field strength will be

Ex=2cm=-ddxVx=2cm=-0-103-1=5V/m

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 91Ó°ÊÓ!

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 potential along the x-axis is V=100e-2xV, wherex is in meters. What is Ex at (a)role="math" localid="1649581712352" x=1.0m and (b) role="math" localid="1649581719374" x=2.0m?

High-frequency signals are often transmitted along a coaxial cable, such as the one shown in FIGURE P26.66. For example, the cable TV hookup coming into your home is a coaxial cable. The signal is carried on a wire of radius R1while the outer conductor of radius R2is grounded (i.e., at V=0V). An insulating material fills the space between them, and an insulating plastic coating goes around the outside.

a. Find an expression for the capacitance per meter of a coaxial cable. Assume that the insulating material between the cylinders is air.

b. Evaluate the capacitance per meter of a cable having R1=0.50mmand R2=3.0mm.

a. Find an expression for the capacitance of a spherical capacitor, consisting of concentric spherical shells of radii R1(inner shell) and R2(outer shell).

b. A spherical capacitor with a 1.0mmgap between the shells has a capacitance of 100pF. What are the diameters of the two spheres?

Two 2.0cm×2.0cmmetal electrodes are spaced 1.0mmapart and connected by wires to the terminals of a 9.0Vbattery.

a. What are the charge on each electrode and the potential difference between them? While the plates are still connected to the battery, insulated handles are used to pull them apart to a new spacing of 2.0mm.

b. What are the charge on each electrode and the potential difference between them?

The two metal spheres in Figure Q26.9 are connected by a metal wire with a switch in the middle. Initially the switch is open. Sphere 1, with the larger radius, is given a positive charge. Sphere 2, with the smaller radius, is neutral. Then the switch is closed. Afterward, sphere 1 has charge Q1, is at potential V1, and the electric field strength at its surface is E1. The values for sphere 2 are Q2,V2,andE2.

a. Is V1larger than, smaller than, or equal to V2? Explain.

b. Is Q1larger than, smaller than, or equal to Q2? Explain.

c. Is E1 larger than, smaller than, or equal to E2? Explain.

See all solutions

Recommended explanations on Physics Textbooks

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.