/*! 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 15 The charges and coordinates of t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The charges and coordinates of two charged particles held fixed in an \(x y\) plane are \(q_{1}=+3.0 \mu \mathrm{C}, x_{1}=3.5 \mathrm{~cm}, y_{1}=0.50 \mathrm{~cm}\), and \(q_{2}=-4.0 \mu \mathrm{C}, x_{2}=-2.0 \mathrm{~cm}, y_{2}=1.5 \mathrm{~cm}\). Find the (a) magni- tude and (b) direction of the electrostatic force on particle 2 due to particle 1 . At what (c) \(x\) and (d) \(y\) coordinates should a third particle of charge \(q_{3}=+4.0 \mu \mathrm{C}\) be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

Short Answer

Expert verified
The force is 34.77 N at 169.69°; place a charge at (-8.43 cm, 1.5 cm).

Step by step solution

01

Calculate the distance between the two charges

First, we need to find the distance between the two charged particles. Use the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Substituting the given values, we get:\[d = \sqrt{((-2.0) - 3.5)^2 + (1.5 - 0.5)^2} = \sqrt{(-5.5)^2 + (1.0)^2} = \sqrt{30.25 + 1.0} \approx 5.57 \, \text{cm}\]
02

Calculate the magnitude of the electrostatic force

To find the magnitude of the force, use Coulomb’s Law: \( F = \frac{k |q_1 q_2|}{d^2} \), where \(k = 8.99 \times 10^9 \frac{\text{N m}^2}{\text{C}^2}\).\[F = \frac{8.99 \times 10^9 \times |3.0 \times 10^{-6} \times (-4.0 \times 10^{-6})|}{(0.0557)^2}\]This simplifies to:\[F \approx \frac{8.99 \times 10^9 \times 12 \times 10^{-12}}{0.0031} \approx 34.77 \, \text{N}\]
03

Determine the direction of the force

The direction of the force on particle 2 due to particle 1 is toward particle 1 because one charge is positive and the other is negative, thus attracting each other. The angle \(\theta\) with respect to the positive \(x\)-axis can be calculated using:\[ \theta = \tan^{-1}\left(\frac{y_2 - y_1}{x_2 - x_1}\right) = \tan^{-1}\left(\frac{1.5 - 0.5}{-2.0 - 3.5}\right) \]This simplifies to:\[\theta = \tan^{-1}\left(\frac{1.0}{-5.5}\right) \approx -10.31^\circ\]Since the force is directed towards particle 1, the actual angle in the coordinate plane is \(-10.31^\circ + 180^\circ = 169.69^\circ\).
04

Set up the condition for the zero net force on particle 2

To cancel the force due to particle 1, particle 3 must exert an equal and opposite force on particle 2. If the force due to particle 3 is \( F_{23} \) equal in magnitude to \( F \) but in the opposite direction, we have:\[ F_{23} = F \, \Rightarrow \, \frac{k |q_2 q_3|}{r_{23}^2} = 34.77 \]
05

Solve for the position of the third charge

To find the coordinates \((x_3, y_3)\), let's assume it lies somewhere on the line connecting particles 1 and 2, extending beyond particle 2. Use:\[ r_{23}^2 = \frac{k |q_2 q_3|}{F} = \frac{8.99 \times 10^9 \times 4.0 \times 10^{-6} \times 4.0 \times 10^{-6}}{34.77}\]Calculate:\[r_{23}^2 = \frac{143.84 \times 10^{-6}}{34.77} \approx 0.004132 \, \text{m}^2\]\[r_{23} \approx 0.0643 \, \text{m} \approx 6.43 \, \text{cm}\]Given the symmetry and arrangement, place the charge along the line, continuing from \((-2.0, 1.5)\). Position at approximately \((x_3, y_3)\), where it negates \(F\). After detailed calculations: \(x_3 = -2.0 \, \text{cm} - 6.43 \, \text{cm} = -8.43 \, \text{cm}\) and \(y_3 = 1.5 \, \text{cm}\).

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.

Coulomb's Law
In electrostatics, understanding how charged particles interact is key to solving problems related to forces and motion. One of the foundational principles governing these interactions is Coulomb's Law. This law describes the force between two point charges. The formula is given by:
  • \( F = \frac{k |q_1 q_2|}{d^2} \)
Where:
  • \( F \) is the magnitude of the electrostatic force.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \frac{\text{N m}^2}{\text{C}^2} \).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( d \) is the distance between the centers of the two charges.
Coulomb's Law tells us that the force is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them. This means if you double the distance, the force becomes one-fourth as strong. Using this principle allows us to calculate how strong the force will be and how it will affect each charge. Forces can be attractive (between opposite charges) or repulsive (between like charges), which impacts the direction of the force vector.
Electrostatic Force Calculation
To calculate the electrostatic force between two charges, you first need to determine the distance between them, which can be done using the coordinate distance formula:
  • \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Once you have the distance, plug it into Coulomb's Law to find the force. In the given exercise, you have two charges: \( q_1 = +3.0 \mu C \) and \( q_2 = -4.0 \mu C \), with coordinates for each particle. After calculating the distance as approximately 5.57 cm, you apply Coulomb's Law:
  • \( F = \frac{8.99 \times 10^9 \times |3.0 \times 10^{-6} \, \text{C} \times (-4.0 \times 10^{-6} \, \text{C})|}{(0.0557)^2} \)
After solving the equation, the magnitude of the force comes out to about 34.77 N. Remember, since the forces are vector quantities, they have both magnitude and direction, with the direction determined by the nature of the charges (attraction in this case because one charge is positive and the other negative). This direction can be expressed in terms of an angle relative to a reference axis, often requiring some trigonometric calculations to resolve fully.
Coordinate Geometry in Physics
Coordinate geometry, also known as analytic geometry, plays a vital role in understanding and solving physics problems involving spatial relationships, such as distances and angles between points. In physics, especially in electrostatic problems, we frequently need to calculate relative positions and directions of forces between particles.The basic tool we use is the distance formula derived from the Pythagorean theorem:
  • \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
This formula allows us to find the straight-line distance between any two points in a plane. Knowing this distance is essential for applying Coulomb’s Law when charges are not placed along a straight line.In the presented exercise, coordinate geometry aids in not only calculating the distance between charges but also in determining angles for direction using trigonometric functions. For instance, the angle \( \theta \) with respect to the positive \( x \)-axis is found using the arctangent function:
  • \( \theta = \tan^{-1}\left(\frac{y_2 - y_1}{x_2 - x_1}\right) \)
Such calculations are crucial in ensuring that the computed direction aligns correctly with the problem's geometry. Understanding these concepts ensures a comprehensive approach to solving physics problems involving multiple dimensions.

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

A charged nonconducting rod, with a length of \(2.00 \mathrm{~m}\) and a cross- sectional area of \(4.00 \mathrm{~cm}^{2}\), lies along the positive side of an \(x\) axis with one end at the origin. The volume charge density \(\rho\) is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if \(\rho\) is (a) uniform, with a value of \(-4.00 \mu \mathrm{C} / \mathrm{m}^{3}\), and (b) nonuniform, with a value given by \(\rho=b x^{2}\), where \(b=-2.00 \mu \mathrm{C} / \mathrm{m}^{5}\) ?

A nonconducting spherical shell, with an inner radius of \(4.0 \mathrm{~cm}\) and an outer radius of \(6.0 \mathrm{~cm}\), has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density \(\rho\) is the charge per unit volume, with the unit coulomb per cubic meter. For this shell \(\rho=b / r\), where \(r\) is the distance in meters from the center of the shell and \(b=3.0 \mu \mathrm{C} / \mathrm{m}^{2} .\) What is the net charge in the shell?

What is the magnitude of the electrostatic force between a singly charged sodium ion \(\left(\mathrm{Na}^{+}\right.\), of charge \(+e\) ) and an adjacent singly charged chlorine ion \(\left(\mathrm{Cl}^{-}\right.\), of charge \(\left.-e\right)\) in a salt crystal if their separation is \(2.82 \times 10^{-10} \mathrm{~m}\) ?

Figure \(21-37\) shows four identical conducting spheres that are actually well separated from one another. Sphere \(W\) (with an initial charge of zero) is touched to sphere \(A\) and then they are separated. Next, sphere \(W\) is touched to sphere \(B\) (with an initial charge of \(-32 e\) ) and then they are separated. Finally, sphere \(W\) is touched to sphere \(C\) (with an initial charge of \(+48 e\) ), and then they are separated. The final charge on sphere \(W\) is \(+18 e\). What was the initial charge on sphere \(A\) ?

An electron is in a vacuum near Earth's surface and located at \(y=0\) on a vertical \(y\) axis. At what value of \(y\) should a second electron be placed such that its electrostatic force on the first electron balances the gravitational force on the first electron?

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.