/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A charge \(+q\) is located at th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A charge \(+q\) is located at the origin, while an identical charge is located on the \(x\) axis at \(x=+0.50 \mathrm{m}\). A third charge of \(+2 q\) is located on the \(x\) axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

Short Answer

Expert verified
Place the third charge at x ≈ 0.707 m on the x-axis.

Step by step solution

01

Understanding the Problem

We need to find the position of the third charge on the x-axis such that it doubles the force experienced by the charge at the origin, without changing the direction of this force. We have two positive charges initially, one at the origin and one at x = 0.50 m.
02

Calculating Initial Force on Charge at Origin

Use Coulomb's Law to calculate the force on the charge at the origin due to the second charge at x = 0.50 m. The formula is \[ F = k_e \frac{q_1 q_2}{r^2} \]where \(k_e = 8.99 \times 10^9 \, \mathrm{N\cdot m^2/C^2}\) is Coulomb's constant, and both charges are \(+q\). The distance \(r\) is 0.50 m.
03

Expressing the Force in Terms of Algebraic Expressions

Substitute the values into the formula:\[ F_\text{origin, initial} = k_e \frac{q \times q}{(0.50)^2} = k_e \frac{q^2}{0.25} = 4k_e q^2 \]This is the initial electrostatic force acting on the charge at the origin due to the charge at x = 0.50 m.
04

Doubling the Force with the Third Charge

Adding the third charge \(+2q\) on the x-axis such that the total force on the charge at the origin is doubled. The direction remains unchanged, which means the third charge must be placed between the origin and the charge at x = 0.50 m, so that the force it exerts on the origin is also repulsive.
05

Setting Up the Equation for Double Force

The third charge should satisfy:\[ 2 \times F_\text{origin, initial} = F_\text{origin, second} + F_\text{origin, third} \]where\[ F_\text{origin, third} = k_e \frac{q \times 2q}{(x^2)} = \frac{2k_e q^2}{x^2} \]Thus:\[ 2 \times 4k_e q^2 = 4k_e q^2 + \frac{2k_e q^2}{x^2} \]
06

Solving for x

Simplifying:\[ 8k_e q^2 = 4k_e q^2 + \frac{2k_e q^2}{x^2} \]Subtracting \(4k_eq^2\) from both sides:\[ 4k_e q^2 = \frac{2k_e q^2}{x^2} \]Dividing both sides by \(k_e q^2\) leaves:\[ 4 = \frac{2}{x^2} \]Solving for \(x^2\):\[ x^2 = \frac{2}{4} = 0.5 \]Taking the square root, we find:\[ x = 0.707 \text{ m} \]
07

Interpreting the Result

The third charge \(+2q\) should be placed at approximately \(x = 0.707\, \text{ m}\) on the x-axis to double the force on the charge at the origin while keeping the direction unchanged.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law is fundamental in understanding the forces between charged objects. It states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of their charges and inversely proportional to the square of the distance between them. The formula is given by:\[F = k_e \frac{q_1 q_2}{r^2}\]where:
  • \(F\) is the force between the charges,
  • \(k_e = 8.99 \times 10^9 \, \mathrm{N\cdot m^2/C^2}\) is Coulomb’s constant,
  • \(q_1\) and \(q_2\) are the magnitudes of the charges,
  • \(r\) is the distance between the charges.
This law helps us understand that forces are stronger when charges are closer and when their magnitudes are larger. It's important to remember that these forces are vector quantities, meaning they have both magnitude and direction.
Electric Charge Placement
In electrostatics, the placement of charges is crucial for determining the forces experienced by these charges. In the given problem, charges are placed along the x-axis:
  • A charge \(+q\) at the origin,
  • Another identical charge \(+q\) at \(x = 0.50\, \text{m}\),
  • A third charge \(+2q\) that needs to be positioned such that its effect is carefully calculated.
For calculations, the relative positioning affects how the forces from each charge interact or add up. In this scenario, the third charge needs to be placed such that its additional force doubles the net force on the origin charge without altering its direction. This requires positioning the \(+2q\) charge somewhere along the positive x-axis but closer to the origin than the \(+q\) charge is.
Net Electrostatic Force
The net electrostatic force is the total force acting on a charge due to the presence of other charges. It's a vector sum of the individual forces exerted by each charge. In our problem, the initial force is calculated using one charge at \(x = 0.50\, \text{m}\) on the charge at the origin, resulting in:\[F_{\text{origin, initial}} = 4k_e q^2\]To double this force with a third charge \(+2q\), calculations involve setting:\[2 \times F_{\text{origin, initial}} = F_{\text{origin, second}} + F_{\text{origin, third}}\]The purpose of setting this equation is to understand how the third charge contributes to the total force, ensuring the same direction by adding forces algebraically, which in this case both aid in repulsive interactions along the x-axis.
Physics Problem Solving
Solving physics problems effectively involves breaking the problem down into manageable parts:
  • Understanding the Problem: Clearly identify the knowns and unknowns. Here, we need the position of the third charge so the net force doubles, and its direction remains unchecked.

  • Applying Relevant Equations: Use equations like Coulomb's Law to frame the problem mathematically according to what is needed.

  • Algebraic Manipulation: Carefully manipulate algebraic equations to isolate variables and solve for unknowns, as seen when determining the third charge's position through solving for \(x\).

  • Interpreting Results: Finally, evaluate the solution contextually to see if it makes physical sense, such as the placement of \(+2q\) at \(x = 0.707 \, \text{m}\) to satisfy all conditions of the problem.
This structured approach ensures that each aspect of the physics problem is comprehensively addressed, leading to a correct and meaningful solution.

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

ssm Iron atoms have been detected in the sun's outer atmosphere, some with many of their electrons stripped away. What is the net electric charge (in coulombs) of an iron atom with 26 protons and 7 electrons? Be sure to include the algebraic sign \((+\) or \(-\) ) in your answer.

Four identical metallic objects carry the following charges: +1.6 \(+6.2,-4.8,\) and \(-9.4 \mu \mathrm{C} .\) The objects are brought simultaneously into contact, so that each touches the others. Then they are separated. (a) What is the final charge on each object? (b) How many electrons (or protons) make up the final charge on each object?

ssm At a distance \(r_{1}\) from a point charge, the magnitude of the electric field created by the charge is \(248 \mathrm{N} / \mathrm{C}\). At a distance \(r_{2}\) from the charge, the field has a magnitude of \(132 \mathrm{N} / \mathrm{C}\). Find the ratio \(r_{2} / r_{1}\).

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is \(\sigma=1.8 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2},\) and the plate separation is \(1.5 \times 10^{-2} \mathrm{m} .\) How fast is the electron moving just before it reaches the positive plate?

A circular surface with a radius of \(0.057 \mathrm{m}\) is exposed to a uniform external electric field of magnitude \(1.44 \times 10^{4} \mathrm{N} / \mathrm{C}\). The magnitude of the electric flux through the surface is \(78 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\). What is the angle (less than \(90^{\circ}\) ) between the direction of the electric field and the normal to the surface?

See all solutions

Recommended explanations on Physics Textbooks

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.