Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 72

An open-ended cylinder of radius \(a\) and length \(2 a\) carries charge \(q\) spread uniformly over its surface. Find the potential at the center of the cylinder. (Hint: Treat the cylinder as a stack of charged rings, and integrate.)

Short Answer

Expert verified
The potential at the center of the cylinder is given by the evaluated integral, which when substituted for \(\lambda\) will give the potential in terms of \(q, a\), and \(\epsilon_0\).

Step by step solution

01

Determine the charge density

Given that the charge \(q\) is uniformly spread over the surface of the cylinder, which has an area \(2 \pi a^2 + 2 \pi a(2a)\). The linear charge density \(\lambda\) can be found by dividing the total charge by the total surface area of the cylinder. \[\lambda = \frac{q}{2 \pi a^2 + 2 \pi a(2a)}\]
02

Break down the cylinder into charged rings

Imagine slicing the cylinder into infinitesimally thin rings. Each ring has a width \(dx\), a radius \(a\), and is at a distance \(x\) from the center of the cylinder. The charge \(dq\) on each ring would therefore be:\[dq = \lambda (2 \pi a dx)\]
03

Calculate the potential by integration

The expression for the electric potential at the center due to a infinitesimally thin ring of charge at a distance \(x\) is given by \(V = \frac{dq}{4 \pi \epsilon_0 x}\). Integrate this from \(-a\) to \(a\) to obtain the potential at the center due to all the ring slices (or the entire cylinder).\[V = \int_{-a}^{a} \frac{\lambda (2 \pi a dx)}{4 \pi \epsilon_0 x}\]Solve this integral for \(V\).
04

Substitute the value for the charge density

Substitute the calculated value of \(\lambda\) from Step 1 into \(V\) to get the potential at the center of the cylinder in terms of the given quantities \(q, a\), and \(\epsilon_0\).

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.

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 proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a \(100-\mathrm{V}\) potential difference. How much energy does each gain?

Two metal spheres each \(1.0 \mathrm{cm}\) in radius are far apart. One sphere carries 38 nC. the other -10 nC. (a) What's the potential on each? (b) If the spheres are connected by a thin wire, what will be the potential on each once equilibrium is reached? (c) How much charge moves between the spheres in order to achieve equilibrium?

Two points \(A\) and \(B\) lie \(15 \mathrm{cm}\) apart in a uniform electric field, with the path \(A B\) parallel to the field. If the potential difference \(\Delta V_{A B}\) is \(840 \mathrm{V},\) what's the field strength?

Proton-beam therapy is preferable to X rays for cancer treatment because protons deliver most of their energy to the tumor, with less damage to healthy tissue. A cyclotron used to accelerate protons for cancer treatment repeatedly passes the protons through a 15-kV potential difference. (a) How many passes are needed to bring the protons' kinetic energy to \(1.2 \times 10^{-11} \mathrm{J} ?\) (b) What's the resulting proton energy in electronvolts?

You're sizing a new electric transmission line, and you can save money with thinner wire. The potential difference between the line and the ground, \(60 \mathrm{m}\) below, is \(115 \mathrm{kV}\). The field at the wire surface cannot exceed \(25 \%\) of the 3 -MV/m breakdown field in air. Neglecting charges in the ground itself, what minimum wire diameter do you specify? (Hint. You'll have to do a numerical calculation.)
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Linear Momentum

Read Explanation

Waves Physics

Read Explanation

Quantum Physics

Read Explanation

Modern Physics

Read Explanation

Physical Quantities And Units

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