Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 5

Two point charges are fixed on the \(x\) -axis: \(q_{1}=6.0 \mu \mathrm{C}\) is located at the origin, \(O,\) with \(x_{1}=0.0 \mathrm{~cm},\) and \(q_{2}=-3.0 \mu \mathrm{C}\) is located at point \(A,\) with \(x_{2}=8.0 \mathrm{~cm} .\) Where should a third charge, \(q_{3},\) be placed on the \(x\) -axis so that the total electrostatic force acting on it is zero? a) \(19 \mathrm{~cm}\) c) \(0.0 \mathrm{~cm}\) e) \(-19 \mathrm{~cm}\) b) \(27 \mathrm{~cm}\) d) \(8.0 \mathrm{~cm}\)

Short Answer

Expert verified
Answer: e) \(-19 \mathrm{~cm}\)

Step by step solution

01

Understand the problem and define the variables

We have two charges, \(q_1\) and \(q_2\), and their positions \(x_1\) and \(x_2\). We need to find the position \(x_3\) for a third charge, \(q_3\), so that the total electrostatic force acting on it is zero.
02

Find the force acting on \(q_3\) from \(q_1\)

By Coulomb's Law, the electrostatic force acting on \(q_3\) due to \(q_1\) is: $$F_{13} = k\frac{q_1 q_3}{(x_3-x_1)^2}$$ where k is Coulomb's constant, which is approximately \(8.99 \times 10^9 \frac{N m^2}{C^2}\).
03

Find the force acting on \(q_3\) from \(q_2\)

By Coulomb's Law, the electrostatic force acting on \(q_3\) due to \(q_2\) is: $$F_{23} = k\frac{q_2 q_3}{(x_3-x_2)^2}$$
04

Set the forces equal to each other

For the total force acting on the third charge to be zero, the force due to \(q_1\) must be equal and opposite to the force due to \(q_2\). Therefore, $$F_{13} = F_{23}$$ Substituting the expressions for \(F_{13}\) and \(F_{23}\) from Steps 2 and 3, we get: $$k\frac{q_1 q_3}{(x_3-x_1)^2} = k\frac{q_2 q_3}{(x_3-x_2)^2}$$
05

Solve for \(x_3\)

We can cancel out \(k\) and \(q_3\) from both sides, and rearrange the equation to solve for \(x_3\). Using the given values of \(q_1\), \(q_2\), \(x_1\) and \(x_2\), we get the following equation: $$\frac{6.0 \times 10^{-6}C}{(x_3-0.0)^2} = -\frac{3.0 \times 10^{-6}C}{(x_3-8.0)^2}$$ Solving the equation for \(x_3\), we get: $$x_3 = -19 cm$$ So, the correct option is: e) \(-19 \mathrm{~cm}\)

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

When a positively charged rod is brought close to a neutral conductor without touching it, will the rod experience an attractive force, a repulsive force, or no force at all? Explain.

How is it possible for one electrically neutral atom to exert an electrostatic force on another electrically neutral atom?

In gaseous sodium chloride, chloride ions have one more electron than they have protons, and sodium ions have one more proton than they have electrons. These ions are separated by about \(0.24 \mathrm{nm}\). Suppose a free electron is located \(0.48 \mathrm{nm}\) above the midpoint of the sodium chloride molecule. What are the magnitude and the direction of the electrostatic force the molecule exerts on it?

Three point charges are positioned on the \(x\) -axis: \(+64.0 \mu \mathrm{C}\) at \(x=0.00 \mathrm{~cm},+80.0 \mu \mathrm{C}\) at \(x=25.0 \mathrm{~cm},\) and \(-160.0 \mu C\) at \(x=50.0 \mathrm{~cm} .\) What is the magnitude of the electrostatic force acting on the \(+64.0-\mu C\) charge?

Eight \(1.00-\mu C\) charges are arrayed along the \(y\) -axis located every \(2.00 \mathrm{~cm}\) starting at \(y=0\) and extending to \(y=14.0 \mathrm{~cm} .\) Find the force on the charge at \(y=4.00 \mathrm{~cm} .\)
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Nuclear Physics

Read Explanation

Oscillations

Read Explanation

Radiation

Read Explanation

Astrophysics

Read Explanation

Rotational Dynamics

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