/*! 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 68 When point charges \(q_{1}=+8.4 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When point charges \(q_{1}=+8.4 \mu \mathrm{C}\) and \(q_{2}=+5.6 \mu \mathrm{C}\) are brought near each other, each experiences a repulsive force of magnitude \(0.66 \mathrm{N}\). Determine the distance between the charges.

Short Answer

Expert verified
The distance between the charges is approximately 0.799 m or 79.9 cm.

Step by step solution

01

Identify the formula

Find the relevant formula to solve the problem. Since we are dealing with point charges and electrostatic force, we use Coulomb's Law, which is given by: \[ F = k \frac{|q_1 q_2|}{r^2} \]where:- **F** is the force between the charges,- **k** is Coulomb's constant \(8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2\),- \(q_1\) and \(q_2\) are the magnitudes of the charges,- **r** is the distance between the centers of the two charges.
02

Rearrange the formula

Re-arrange the formula to solve for the distance **r** between the charges:\[ r = \sqrt{\frac{k |q_1 q_2|}{F}} \]
03

Substitute the given values

Substitute the given values into the rearranged formula:- \( F = 0.66 \mathrm{N}\),- \( q_1 = 8.4 \times 10^{-6} \mathrm{C} \) (convert \(\mu C\) to \(C\)),- \( q_2 = 5.6 \times 10^{-6} \mathrm{C} \).The formula becomes:\[ r = \sqrt{\frac{8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2 \times (8.4 \times 10^{-6} \mathrm{C}) \times (5.6 \times 10^{-6} \mathrm{C})}{0.66 \mathrm{N}}} \]
04

Perform the calculations

Calculate the expression inside the square root first:\[ 8.99 \times 10^9 \times 8.4 \times 10^{-6} \times 5.6 \times 10^{-6} \approx 4.218816 \times 10^{-1} \]Then, divide by the force:\[ \frac{4.218816 \times 10^{-1}}{0.66} \approx 6.390931 \times 10^{-1} \]Now take the square root:\[ \sqrt{6.390931 \times 10^{-1}} \approx 0.799432 \]
05

State the distance

Since our calculations give the distance in standard meters (m), the final result is that the distance between the two charges is approximately:\[ r \approx 0.799 \mathrm{m} \] or 79.9 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.

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.

Electrostatic Force Calculation
Electrostatic force refers to the force between charged objects due to their electric fields. When dealing with such forces, we often use Coulomb's Law to calculate the magnitude of the force. The formula provided by Coulomb's Law is:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:
  • \(F\) is the electrostatic force between two point charges.
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \ \mathrm{Nm^2/C^2}\).
  • \(q_1\) and \(q_2\) are the magnitudes of the charges in coulombs.
  • \(r\) is the distance between the centers of the two charges in meters.
To find the force, you plug the values of the charges, distance, and Coulomb's constant into the formula. In an example scenario with two specific charges, you'll have the force calculated directly by adjusting for given variables, as demonstrated in the step-by-step solution. This formula highlights the inverse square relationship between the force and distance, making distance a crucial factor in determining the magnitude of the electrostatic force.
Point Charges Interaction
Point charges are idealized objects that possess a charge concentrated at a single point in space, allowing simplifications in calculations. In reality, charged objects have a finite size, but treating them as point charges for calculations is practical when they're small enough compared to the distance between them.

The interaction between point charges depends on the nature of their charges: like charges (both positive or both negative) will repel each other, whereas opposite charges will attract. This interaction is quantified through the electrostatic force calculated using Coulomb’s Law. In our specific example, both charges are positive, leading to a repulsive force, an expected scenario reflected in the result of the given magnitude of force.
  • Repulsion occurs between like charges.
  • Attraction occurs between opposite charges.
Understanding how charges interact helps in predicting the resulting movements of charged particles in various fields, from electric circuits to natural phenomena.
Distance Between Charges
The distance between two interacting point charges is a significant factor in determining the strength of the electrostatic force. As Coulomb's Law indicates, the force is inversely proportional to the square of the distance. This means the force decreases dramatically as the distance increases.

When solving problems involving point charges, one may need to rearrange Coulomb’s Law to solve for distance, especially when the force is known, as shown in the solution to our exercise:\[ r = \sqrt{\frac{k |q_1 q_2|}{F}} \]Here, knowing the magnitude of the force, charges, and Coulomb's constant allows for the calculation of the distance \(r\). Such calculations are crucial not only for theoretical problems but also in applications across various technological fields, facilitating the design of precise electronic instruments and ensuring safety in handling charged materials.
Understanding this relationship empowers students to manipulate variables adeptly in physics, achieving insights into the balance forces exert at different scales.

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 objects carry initial charges that are \(q_{1}\) and \(q_{2}\), respectively, where \(\left|q_{2}\right|>\left|q_{1}\right| .\) They are located \(0.200 \mathrm{m}\) apart and behave like point charges. They attract each other with a force that has a magnitude of \(1.20 \mathrm{N} .\) The objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to the magnitude of the initial attractive force. What are the magnitudes of the initial charges on the objects? Section 18.6 The Electric Field Section 18.7 Electric Field Lines Section 18.8 The Electric Field Inside a Conductor: Shielding

A charge \(Q\) is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: \(\Phi_{1}=+1500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) \(\Phi_{2}=+2200 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{3}=+4600 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{4}=-1800 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\) \(\Phi_{5}=-3500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) and \(\Phi_{6}=-5400 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} .\) What is \(Q ?\)

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?

A particle of charge \(+12 \mu C\) and \(\operatorname{mass} 3.8 \times 10^{-5} \mathrm{kg}\) is released from rest in a region where there is a constant electric field of \(+480 \mathrm{N} / \mathrm{C}\). What is the displacement of the particle after a time of \(1.6 \times 10^{-2} \mathrm{s} ?\)

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?

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.