/*! 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 30 Particles 1 and 2 are fixed in p... [FREE SOLUTION] | 91Ó°ÊÓ

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Particles 1 and 2 are fixed in place on an \(x\) axis, at a separation of \(L=8.00 \mathrm{~cm} .\) Their charges are \(q_{1}=+e\) and \(q_{2}=-27 e\). Particle 3 with charge \(q_{3}=+4 e\) is to be placed on the line between particles 1 and 2 , so that they produce a net electrostatic force \(\vec{F}_{3, \text { net }}\) on it. (a) At what coordinate should particle 3 be placed to minimize the magnitude of that force? (b) What is that minimum magnitude?

Short Answer

Expert verified
(a) Place at 1.6 cm from particle 1. (b) Minimum force is balanced at this point.

Step by step solution

01

Understanding Forces on Particle 3

To minimize the net force on particle 3 positioned between particles 1 and 2, we need to balance the electrostatic forces from both particles on particle 3. The equation for the force between two charges is given by Coulomb's law: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( F \) is the force, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the point charges, and \( r \) is the distance between the charges.
02

Setting Up the Coordinate and Net Force Equation

Let particle 3 be placed at a distance \( x \) from particle 1, which positions it at \( x \) on the axis. The distance from particle 3 to particle 2 will therefore be \( L - x = 8.00 \, \text{cm} - x \). The force that particle 1 exerts on particle 3 is:\[ F_{13} = k \frac{q_1 q_3}{x^2} \]The force that particle 2 exerts on particle 3 is:\[ F_{23} = k \frac{q_2 q_3}{(L-x)^2} \]The equation for the net force on particle 3 is:\[ F_{3, \text{net}} = F_{13} + F_{23} \]
03

Balancing Forces to Minimize Net Force

For minimal net force, the forces exerted by particles 1 and 2 need to cancel each other. This means \[ F_{13} = -F_{23} \]Substituting the forces from Step 2, we get:\[ k \frac{(\+e)(\+4e)}{x^2} = -k \frac{(-27e)(\+4e)}{(L-x)^2} \]Simplifying, we obtain:\[ \frac{4}{x^2} = \frac{108}{(L-x)^2} \]
04

Solving the Equation for x

Simplifying the equation from Step 3:\[ 4(L-x)^2 = 108x^2 \]\[ 4L^2 - 8Lx + 4x^2 = 108x^2 \]\[ 4L^2 - 8Lx - 104x^2 = 0 \]Dividing through by 4:\[ L^2 - 2Lx - 26x^2 = 0 \]Substitute \( L = 8.00 \, \text{cm} \):\[ 64 - 16x - 26x^2 = 0 \]Solve this quadratic equation for \( x \). Reducing terms, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -26 \), \( b = -16 \), and \( c = 64 \).
05

Finding the Minimum Force Magnitude

Solve the quadratic equation for the smallest real positive value of \( x \) as the minimum displacement from particle 1. Upon solving, this occurs at approximately \( x = 1.6 \, \text{cm} \).We substitute \( x = 1.6 \, \text{cm} \) back into one of the force equations to find the magnitude of the force, such as:\[ F_{13} = k \frac{e \cdot 4e}{(0.016)^2} \] (convert \( cm \) to \( m \)), use value of \( k \). Find the equivalent \( F_{23}\) using the same method and confirm balance.

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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 how charged particles interact. It describes the force between two point charges. The formula for this is \[ F = k \frac{|q_1 q_2|}{r^2} \] where:
  • \( F \) is the electrostatic force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance between the centers of the two charges.
This law shows that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Simply put, larger charges or shorter distances result in a stronger force.
Electrostatic Force
Electrostatic force comes up often in physics, especially when discussing electric charges. It's the force acting between charges and can be attractive or repulsive. When two charges are of opposite types, like one positive and one negative, they attract each other. If both charges are of the same type, either both positive or both negative, they repel. In this exercise, we calculated the forces exerted by two fixed charges (particles 1 and 2) on a movable charge (particle 3). Understanding how to calculate and balance these forces is key to solving problems involving more than one charge. Using Coulomb's Law, we determined the forces based on position, placing particle 3 to minimize the net force acting upon it.
Quadratic Equation
Quadratic equations are pivotal in solving certain physics problems like this one, where finding the position that minimizes force requires solving a quadratic formula.The derived equation for this problem was \[ 64 - 16x - 26x^2 = 0 \] which represents a standard quadratic form \( ax^2 + bx + c = 0 \).To solve it, use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where:
  • \( a = -26 \)
  • \( b = -16 \)
  • \( c = 64 \)
Plugging in these values gives two possible solutions for \( x \), with the realistic one (non-negative in physical terms) indicating the position for minimal force.
Charge Interaction
Charge interaction is a straightforward yet crucial concept in electrostatics. It deals with the way electric charges exert forces on each other.In this scenario, three particles are interacting: two fixed and one movable. The charges' interactions depend heavily on their magnitudes and signs:
  • Particle 1 (\(+e\)) attracts particle 3 (\(+4e\)) due to their charges being opposite, similar attraction occurs with particle 2 (\(-27e\)) and particle 3.
  • These forces occur along the line between the fixed particles, impacting where particle 3 can be placed to experience the least net force.
By correctly balancing these interactions, we can strategically position particle 3 to achieve an equilibrium of forces, demonstrating a practical application of electrostatic principles.

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Most popular questions from this chapter

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?

Particle of charge \(+4 e\) is above a floor by distance \(d_{1}=2.00 \mathrm{~mm}\) and particle 2 of charge \(+6 e\) is on the floor, at distance \(d_{2}=6.00 \mathrm{~mm}\) horizontally from particle \(1 .\) What is the \(x\) component of the electrostatic force on particle 2 due to particle \(1 ?\)

Four particles are fixed along an \(x\) axis, separated by distances \(d=2.00 \mathrm{~cm} .\) The charges are \(q_{1}=+2 e, q_{2}=-e, q_{3}=\) \(+e\), and \(q_{4}=+4 e\), with \(e=1.60 \times 10^{-19} \mathrm{C}\). In unit-vector notation, what is the net electrostatic force on (a) particle 1 and (b) particle 2 due to the other particles?

Two tiny conducting balls of identical mass \(m\) and identical charge \(q\) hang from nonconducting threads of length \(L\). Assume that \(\theta\) is so small that \(\tan \theta\) can be replaced by its approximate equal, \(\sin \theta\). (a) Show that $$ x=\left(\frac{q^{2} L}{2 \pi \varepsilon_{0} m g}\right)^{1 / 3} $$ gives the equilibrium separation \(x\) of the balls. (b) If \(L=120 \mathrm{~cm}, m=10 \mathrm{~g}\), and \(x=5.0 \mathrm{~cm}\), what is \(|q| ?\)

A particle of charge \(+3.00 \times 10^{-6} \mathrm{C}\) is \(12.0 \mathrm{~cm}\) distant from a second particle of charge \(-1.50 \times 10^{-6} \mathrm{C} .\) Calculate the magnitude of the electrostatic force between the particles.

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