Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 26: Q. 4 (page 737)

Figure EX26.4 is a graph of Ex. The potential at the origin is -50V. What is the potential at x=3.0m?

Short Answer

Expert verified

The required potential is-550V

Step by step solution

01

Given information and formula used 

Given :

The potential at the origin : -50V

To find : The potential at :x=3.0m

Graph -

Theory used :

Potential difference is :∆V=Vf-Vi=∫Ex·∆x

02

Calculating the required the potential difference  

Using the formula to calculate potential difference, we have :

∆V=Vf-Vi=∫Ex·∆x⇒V3-V0=-∫02Ex·dx-∫23E·dx⇒V3+50=-200×2-∫23(600-200x)dx↤{fromgraph}⇒V3=-400-50-[600x-200x²2]23 =-450-(900-800)=-450-100=-550V

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A capacitor being charged has a current carrying charge to and away from the plates. In the next chapter we will define current to be dQ/dt the rate of charge flow. What is the current to a 10μF capacitor whose voltage is increasing at the rate of 2.0V/s?

Engineers discover that the electric potential between two electrodes can be modeled as V(x)=V0In(1+x/d), where V0is a constant, xis the distance from the first electrode in the direction of the second, and dis the distance between the electrodes. What is the electric field strength midway between the electrodes?

The electric potential in a region of space is V=200/2x2+y2,where x and y are in meters. What are the strength and direction of the electric field at 1x, y2=12.0 m, 1.0 m2? Give the direction as an angle cw or ccw (specify which) from the positive x-axis

a. Use the methods of Chapter 25 to find the potential at distance xon the axis of the charged rod shown in FIGURE P26.43.

b. Use the result of part a to find the electric field at distance xon the axis of a rod

A nerve cell in its resting state has a membrane potential of -70mV, meaning that the potential inside the cell is 70mV less than the potential outside due to a layer of negative charge on the inner surface of the cell wall and a layer of positive charge on the outer surface. This effectively makes the cell wall a charged capacitor. When the nerve cell fires, sodium ions,Na+, flood through the cell wall to briefly switch the membrane potential to +40mV. Model the central body of a nerve cell-the soma-as a 50μmdiameter sphere with a 7.0-nm-thick cell wall whose dielectric constant is 9.0. Because a cell's diameter is much larger than the wall thickness, it is reasonable to ignore the curvature of the cell and think of it as a parallel-plate capacitor. How many sodium ions enter the cell as it fires?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Physics of Motion

Read Explanation

Geometrical and Physical Optics

Read Explanation

Oscillations

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Atoms and Radioactivity

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.

Sign-up for free