Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 25

A small sphere with a charge of \(-0.60 \mu \mathrm{C}\) is placed in a uniform electric field of magnitude \(1.2 \times 10^{6} \mathrm{N} / \mathrm{C}\) pointing to the west. What is the magnitude and direction of the force on the sphere due to the electric field?

Short Answer

Expert verified
Answer: The magnitude of the force acting on the sphere is 0.72 N, and the direction is to the east.

Step by step solution

01

Identify given quantities

We are given the following information: Charge of the small sphere, \(q = -0.60 \mu \mathrm{C} = -0.60 \times 10^{-6} \mathrm{C}\) Magnitude of the uniform electric field, \(E = 1.2 \times 10^6 \mathrm{N}/\mathrm{C}\)
02

Use the formula for force acting on a charged particle in an electric field

The formula for the force, \(F\), acting on a charged particle in an electric field, \(E\), is given by: \(F = qE\)
03

Calculate the force

Now, we can substitute the given values and calculate the force acting on the charged sphere: \(F = qE = (-0.60 \times 10^{-6} \mathrm{C})(1.2 \times 10^6 \mathrm{N}/\mathrm{C}) = -0.72\,\mathrm{N}\)
04

Determine the direction of the force

Since the charge of the sphere is negative and the electric field is pointing to the west, the force on the sphere due to the electric field will be in the opposite direction, which is to the east.
05

Report the magnitude and direction of the force

The magnitude of the force acting on the sphere is \(0.72\,\mathrm{N}\) in the east direction. However, sometimes the force in the east direction is reported as a positive value, so you could also report the magnitude of the force as \(+0.72\,\mathrm{N}\) to the east.

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 Styrofoam balls with the same mass \(m=9.0 \times 10^{-8} \mathrm{kg}\) and the same positive charge \(Q\) are suspended from the same point by insulating threads of length \(L=0.98 \mathrm{m} .\) The separation of the balls is \(d=0.020 \mathrm{m} .\) What is the charge \(Q ?\)
Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?
A flat conducting sheet of area \(A\) has a charge \(q\) on each surface. (a) What is the electric field inside the sheet? (b) Use Gauss's law to show that the electric field just outside the sheet is \(E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} .\) (c) Does this contradict the result of Problem \(69 ?\) Compare the field line diagrams for the two situations.
Two metal spheres of radius \(5.0 \mathrm{cm}\) carry net charges of $+1.0 \mu \mathrm{C}\( and \)+0.2 \mu \mathrm{C} .$ (a) What (approximately) is the magnitude of the electrical repulsion on either sphere when their centers are \(1.00 \mathrm{m}\) apart? (b) Why cannot Coulomb's law be used to find the force of repulsion when their centers are \(12 \mathrm{cm}\) apart? (c) Would the actual force be larger or smaller than the result of using Coulomb's law with \(r=12 \mathrm{cm} ?\) Explain.
A hollow conducting sphere of radius \(R\) carries a negative charge \(-q .\) (a) Write expressions for the electric field \(\overrightarrow{\mathbf{E}}\) inside \((rR)\) the sphere. Also indicate the direction of the field. (b) Sketch a graph of the field strength as a function of \(r\). [Hint: See Conceptual Example \(16.8 .]\)
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Thermodynamics

Read Explanation

Mechanics and Materials

Read Explanation

Rotational Dynamics

Read Explanation

Fluids

Read Explanation

Electrostatics

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