/*! 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. 16 What is the potential difference... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the potential difference ∆V34in Figure EX26.16?

Short Answer

Expert verified

The potential difference is20V

Step by step solution

01

Given information and theory used   

Given :

Theory used :

To find the potential difference between two points :

Step 1: Determine the strength of the field, E, and the distance between the two points, d.

Step 2: Use the formula ΔV=Edto calculate the potential difference between the two points.

02

Calculating the potential difference 

We may express the potential at point 1 as

∆V41=-60=V1-V4⇒V1=V4-60

by knowing the potential difference between locations 4 and 1.

Again, we may express the potential at point 2 as

∆V12=30=V2-V1⇒V2=V1+30

by knowing the potential difference between locations 1 and 2.

By substituting this result in the previous one, we get

V2=V4-60+30=V4-30

We may express the potential at point 3 as

∆V23=50=V3-V2⇒V3=V2+50

by knowing the potential difference between points 2 and 3.

We may calculate V3=V4-30+50=V4+20by substituting our result for the potential at point 2 at the preceding one.

The possible difference between points 3 and 4 is now obvious:

∆V34=V3-V4=20V.

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

An electric dipole at the origin consists of two charges q spaced apart along the y-axis.

a. Find an expression for the potential V(x, y) at an arbitrary point in the xy-plane. Your answer will be in terms of q, s, x, and y.

b. Use the binomial approximation to simplify your result from part a when s V x and s V y.

c. Assuming s V x and y, find expressions for Ex and Ey, the components of E u for a dipole.

d. What is the on-axis field E? Does your result agree with Equation 23.10?

e. What is the field E u on the bisecting axis? Does your result agree with Equation 23.11?

Initially, the switch in FIGUREP26.61is in position A and capacitors C2and C3are uncharged. Then the switch is flipped to position B. Afterward, what are the charge on and the potential difference across each capacitor?

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.

You need a capacitance of 50mF, but you don’t happen to have a 50mFcapacitor. You do have a 75mFcapacitor. What additional capacitor do you need to produce a total capacitance of 50mF? Should you join the two capacitors in parallel or in series?

Figure Q26.10 shows a 3Vbattery with metal wires attached to each end. What are the potential differences ∆V12=V2-V1,∆V23=V3-V2,∆V34=V4-V3,and∆V41=V1-V4?

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.