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

Another unit of charge is the electrostatic unit (esu). It is defined as follows: Two point charges, each of 1 esu and separated by \(1 \mathrm{~cm}\), exert a force of exactly 1 dyne on each other: 1 dyne \(=1 \mathrm{~g} \mathrm{~cm} / \mathrm{s}^{2}=1 \cdot 10^{-5} \mathrm{~N}\). a) Determine the relationship between the esu and the coulomb. b) Determine the relationship between the esu and the elementarv charoe

The force between a charge of \(25 \mu C\) and a charge of \(-10 \mu C\) is \(8.0 \mathrm{~N}\). What is the separation between the two charges? a) \(0.28 \mathrm{~m}\) c) \(0.45 \mathrm{~m}\) b) \(0.53 \mathrm{~m}\) d) \(0.15 \mathrm{~m}\)

Two equal magnitude negative charges \((-q\) and \(-q)\) are fixed at coordinates \((-d, 0)\) and \((d, 0) .\) A positive charge of the same magnitude, \(q\), and with mass \(m\) is placed at coordinate \((0,0),\) midway between the two negative charges. If the positive charge is moved a distance \(\delta \ll d\) in the positive \(y\) -direction and then released, the resulting motion will be that of a harmonic oscillator-the positive charge will oscillate between coordinates \((0, \delta)\) and \((0,-\delta)\). Find the net force acting on the positive charge when it moves to \((0, \delta)\) and use the binomial expansion \((1+x)^{n} \approx 1+n x,\) for \(x \ll 1,\) to find an expression for the frequency of the resulting oscillation.

In general, astronomical objects are not exactly electrically neutral. Suppose the Earth and the Moon each carry a charge of \(-1.00 \cdot 10^{6} \mathrm{C}\) (this is approximately correct; a more precise value is identified in Chapter 22 ). a) Compare the resulting electrostatic repulsion with the gravitational attraction between the Moon and the Earth. Look up any necessary data. b) What effects does this electrostatic force have on the size, shape, and stability of the Moon's orbit around the Earth?

A current of \(5.00 \mathrm{~mA}\) is enough to make your muscles twitch. Calculate how many electrons flow through your skin if you are exposed to such a current for \(10.0 \mathrm{~s}\).
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Force

Read Explanation

Magnetism

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Thermodynamics

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