Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 23

A point charge, \(q=4.00 \cdot 10^{-9} \mathrm{C},\) is placed on the \(x\) -axis at the origin. What is the electric field produced at \(x=25.0 \mathrm{~cm} ?\)

Short Answer

Expert verified
Answer: The electric field produced by the point charge at a distance of 25.0 cm is 361.9 N/C.

Step by step solution

01

Identify given variables

We are given the following information: - Point charge, \(q = 4.00 \cdot 10^{-9} \mathrm{C}\) - Distance from the point charge, \(x = 25.0 \mathrm{cm}\), which we need to convert to meters: \(x = 0.25 \mathrm{m}\) Our goal is to find the electric field, \(E\), at this distance.
02

Use Coulomb's Law to find the electric field

Coulomb's Law for the electric field produced by a point charge is given by the formula: \(E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\) Where: - \(E\) is the electric field - \(\epsilon_0\) is the vacuum permittivity (electric constant), with a value of \(8.85 \cdot 10^{-12} \mathrm{C^2/N} \cdot \mathrm{m^2}\), - \(q\) is the charge of the point charge - \(r\) is the distance from the point charge to the point where we want to find the electric field. In this exercise, we will use \(x\) instead of \(r\) for distance.
03

Substitute the given values and calculate the electric field

We will now plug in the given values to find the electric field at the desired \(x\): \(E = \frac{1}{4\pi(8.85 \cdot 10^{-12}\,\mathrm{C^2/N} \cdot \mathrm{m^2})} \cdot \frac{4.00 \cdot 10^{-9}\,\mathrm{C}}{(0.25\, \mathrm{m})^2}\) Calculate the electric field: \(E = 361.9\,\mathrm{N/C}\)
04

Write the final answer

The electric field produced by the point charge at \(x = 25.0\,\mathrm{cm}\) is \(361.9\,\mathrm{N/C}\).

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 dipole is completely enclosed by a spherical surface. Describe how the total electric flux through this surface varies with the strength of the dipole.

Many people had been sitting in a car when it was struck by lightning. Why were they able to survive such an experience?

A uniform sphere has a radius \(R\) and a total charge \(+Q,\) uniformly distributed throughout its volume. It is surrounded by a thick spherical shell carrying a total charge \(-Q,\) also uniformly distributed, and having an outer radius of \(2 R\). What is the electric field as a function of \(R ?\)

Two infinite, uniformly charged, flat nonconducting surfaces are mutually perpendicular. One of the surfaces has a charge distribution of \(+30.0 \mathrm{pC} / \mathrm{m}^{2}\), and the other has a charge distribution of \(-40.0 \mathrm{pC} / \mathrm{m}^{2}\). What is the magnitude of the electric field at any point not on either surface?

Consider an electric dipole on the \(x\) -axis and centered at the origin. At a distance \(h\) along the positive \(x\) -axis, the magnitude of electric field due to the electric dipole is given by \(k(2 q d) / h^{3} .\) Find a distance perpendicular to the \(x\) axis and measured from the origin at which the magnitude of the electric field stays the same.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Wave Optics

Read Explanation

Astrophysics

Read Explanation

Physical Quantities And Units

Read Explanation

Medical Physics

Read Explanation

Classical Mechanics

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