Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 8

In considering the potential of an infinite flat sheet, why isn't it useful to take the zero of potential at infinity?

Short Answer

Expert verified
It's not useful to take the zero of potential at infinity for an infinite flat sheet because the electric field it creates is constant everywhere. Thus the potential at infinity is not zero, as the influence of the infinite sheet is never escaped, even at infinity.

Step by step solution

01

Understanding Potential at Infinity

The potential at a point can be defined as the work done by an external force (not the electrostatic force) in bringing a unit positive charge without any acceleration from infinity to that point. In simpler terms, it's how much energy it takes to move a charge to a certain location from infinitely far away. For isolated systems or systems with boundaries, it is safe to assume that the system does not affect the surroundings at infinity, so the potential at infinity is taken as zero.
02

Recognizing the Infinite Flat Sheet

The scenario being dealt with in this exercise is an infinite flat sheet. In theory, an infinite sheet extends indefinitely in all directions on a plane. This means that no matter how far away a point charge is moved, it will still experience a force from the infinite sheet due to electric fields it creates.
03

Reasoning the Potential of an Infinite Flat Sheet

The reason that zero of potential at infinity is not useful for an infinite flat sheet is that the field it creates is constant everywhere. No matter how far a unit positive charge is taken, the electric field due to the infinite sheet doesn't change. Hence, the potential at any point in space isn't zero, even at infinity. Therefore, it doesn't make sense to take the potential at infinity as zero for an infinite flat sheet as we're never 'far enough away' to escape its influence.

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

Two 5.0 -cm-diameter conducting spheres are \(8.0 \mathrm{m}\) apart, and each carries \(0.12 \mu \mathrm{C}\). Determine (a) the potential on each sphere, (b) the field strength at the surface of each sphere, (c) the potential midway between the spheres, and (d) the potential difference between the spheres.

The potential is constant throughout an entire volume. What must be true of the electric field within that volume?

Find the potential as a function of position in the electric field \(\vec{E}=a x \hat{\imath},\) where \(a\) is a constant and \(V=0\) at \(x=0\)

A 2.0 -cm-radius metal sphere carries \(75 \mathrm{nC}\) and is surrounded by a concentric spherical conducting shell of radius \(10 \mathrm{cm}\) carrying -75 nC. (a) Find the potential difference between shell and sphere. (b) How would your answer change if the shell's charge were +150 nC?

Two flat metal plates are a distance \(d\) apart, where \(d\) is small compared with the plate size. If the plates carry surface charge densities \(\pm \sigma,\) show that the magnitude of the potential difference between them is \(V=\sigma d / \epsilon_{0}\)
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Fluids

Read Explanation

Medical Physics

Read Explanation

Electric Charge Field and Potential

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