Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Q13PE (page 664)

Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased?

Short Answer

Expert verified

The separation became 15.times of the original.

Step by step solution

01

Step 1:Given Data

Force increased by a factor of 25

02

Electrostatic force

The force of attraction or repulsion that exists between two-point charges separated by some distance in space is known as electrostatic force.

03

Comparing forces

The electrostatic force between two-point charges q and Q separated by some distance r is given as,

F=KqQr2..........(1.1)

When the separation between the charges is changed to r', the new electrostatic force is given as,

F'=KqQr'2..........(1.2)

Taking the ratio of equations (1.1) and (1.2),

FF'=KqQr'KqQr'2FF'=r'r2

Since F'=25F. Therefore,

F25F=r'r2r'r2=125r'r=15r'=r5

Hence, the separation between the charges became 15of the original.

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

Would the self-created electric field at the end of a pointed conductor, such as a lightning rod, remove positive or negative charge from the conductor? Would the same sign charge be removed from a neutral pointed conductor by the application of a similar externally created electric field? (The answers to both questions have implications for charge transfer utilizing points.)

Find the total Coulomb force on the charge\(q\)in Figure 18.53, given that\(q = {\rm{1}}{\rm{.00 }}\mu {\rm{C}}\),\({q_a} = {\rm{2}}{\rm{.00 }}\mu {\rm{C}}\),\({q_b} = - {\rm{3}}{\rm{.00 }}\mu {\rm{C}}\),\({q_c} = - {\rm{4}}{\rm{.00 }}\mu {\rm{C}}\), and\({q_d} = + {\rm{1}}{\rm{.00 }}\mu {\rm{C}}\). The square is\({\rm{50}}{\rm{.0 cm}}\)on a side.

Find the electric field at the location of \({q_a}\) in Figure 18.53 given that \({q_b} = {q_c} = {q_d} = + 2.00{\rm{ nC}}\), \(q = - 1.00{\rm{ nC}}\), and the square is \({\rm{20}}{\rm{.0 cm}}\) on a side.

(a) Find the electric field at the center of the triangular configuration of charges in Figure 18.54, given that\({q_a} = + {\rm{2}}{\rm{.50 nC}}\),\({q_b} = - {\rm{8}}{\rm{.00 nC}}\), and\({q_c} = + {\rm{1}}{\rm{.50 nC}}\). (b) Is there any combination of charges, other than\({q_a} = {q_b} = {q_c}\), that will produce a zero-strength electric field at the center of the triangular configuration?

Figure 18.54 Point charges located at the corners of an equilateral triangle\(25.0{\rm{ cm}}\)on a side.

(a) Calculate the electric field strength near a 10.0 cm diameter conducting sphere that has 1.00 C of excess charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Measurements

Read Explanation

Physics of Motion

Read Explanation

Torque and Rotational Motion

Read Explanation

Circular Motion and Gravitation

Read Explanation

Force

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