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Chapter 21: Problem 65

Three point charges are arranged along the \(x\) -axis. Charge \(q_{1}=-4.50 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+2.50 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m} .\) A positive point charge \(q_{3}\) is located at the origin. (a) What must the value of \(q_{3}\) be for the net force on this point charge to have magnitude 4.00\(\mu \mathrm{N} ?\) (b) What is the direction of the net force on \(q_{3} ?(\mathrm{c})\) Where along the \(x\) -axis can \(q_{3}\) be placed and the net force on it be zero, other than the trivial answers of \(x=+\infty\) and \(x=-\infty ?\)

Short Answer

Expert verified
(a) Calculate \( q_3 \) using Coulomb's law. (b) Determine force direction by net contribution. (c) Solve for \( x \) where forces cancel.

Step by step solution

01

Understand the problem

We have three point charges, with two of them having fixed positions. We need to determine the value of a third charge for specific force conditions. Charge locations and magnitudes are given, and we need to find how they will interact with each other.
02

Set up the force equation for part (a)

Using Coulomb's Law, find the net force on charge \( q_3 \) due to \( q_1 \) and \( q_2 \). Remember, \( F = k \frac{|q_1 q_3|}{r^2} \), where \( k = 8.99 \times 10^9 \, \text{N m}^2/ ext{C}^2 \). Since \( q_3 \) is positive, calculate the forces separately contributed by \( q_1 \) and \( q_2 \) on \( q_3 \). The net force on \( q_3 \) is the vector sum of these forces.
03

Calculate distances and individual forces

The distance from \( q_3 \) to \( q_1 \) is 0.200 m and to \( q_2 \) is 0.300 m. Let \( F_{31} \) be the force on \( q_3 \) from \( q_1 \) and \( F_{32} \) be the force on \( q_3 \) from \( q_2 \). Use Coulomb's law:\[F_{31} = 8.99 \times 10^9 \frac{|-4.50 \times 10^{-9} q_3|}{(0.200)^2}\quad \text{and}\quad F_{32} = 8.99 \times 10^9 \frac{|2.50 \times 10^{-9} q_3|}{(0.300)^2}\]Calculate these forces.
04

Set up the equation for net force

The net force magnitude is given as \( 4.00 \times 10^{-6} \, \text{N} \). Set up the equation:\[4.00 \times 10^{-6} = \left|F_{31} + F_{32}\right|\]Substitute the expressions for \( F_{31} \) and \( F_{32} \) from Step 3 into this equation to solve for \( q_3 \).
05

Solve for \( q_3 \)

Substitute expressions from Step 3 into Step 4:\[4.00 \times 10^{-6} = \left|8.99 \times 10^9 \left(\frac{-4.50 \times 10^{-9} q_3}{0.200^2} + \frac{2.50 \times 10^{-9} q_3}{0.300^2}\right)\right|\]Solve this equation to find the value of \( q_3 \).
06

Determine the direction of the force

Analyze the direction of the forces \( F_{31} \) and \( F_{32} \), where \( F_{31} \) acts towards \( q_1 \) and \( F_{32} \) acts towards \( q_2 \). Determine the net direction based on the magnitude of these forces.
07

Find the non-trivial position for zero net force

For part (c), find the position \( x \) where the net force on \( q_3 \) is zero. Set up the condition:\[F_{31} = F_{32}\]Use the relation:\[k\frac{|q_1 q_3|}{(x - 0.200)^2} = k\frac{|q_2 q_3|}{(x + 0.300)^2}\]Solve for \( x \) to find the position along the \( x \)-axis where the forces cancel each other.

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Key Concepts

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

Point Charges
Point charges are elementary concepts in physics. They are hypothetical charges located in a single point in space. Despite their simplicity, these charges allow us to model and understand how charges interact with one another under electrostatic conditions.

In the exercise given, three point charges are positioned along the x-axis. These charges are interacting based on their respective magnitudes and positions. Charge locations are given as follows:
  • Charge \( q_1 = -4.50 \, \text{nC} \) located at \( x = 0.200 \, \text{m} \)
  • Charge \( q_2 = +2.50 \, \text{nC} \) located at \( x = -0.300 \, \text{m} \)
  • Charge \( q_3 \) is situated at the origin.
Understanding point charges involves knowing that they exert forces on each other but, in this problem, only on a linear path, simplifying how we calculate those forces.
Electric Force
Electric force is the fundamental interaction between charged objects, causing them to attract or repel each other.

This force is calculated using Coulomb's Law, which is described by the equation:\[F = k \frac{|q_1 q_2|}{r^2}\]where:
  • \( F \) is the magnitude of the force between the charges
  • \( k \) is the Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \)
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges
  • \( r \) is the distance between the charges
In our scenario, we needed to calculate separate forces exerted on charge \( q_3 \) by \( q_1 \) and \( q_2 \). Consider these as:
  • \( F_{31} \) between \( q_3 \) and \( q_1 \)
  • \( F_{32} \) between \( q_3 \) and \( q_2 \)
Evaluating these forces helps to identify how they influence \( q_3 \) and determine its net force.
Net Force Calculation
A net force calculation involves summing up all individual forces acting on an object.

In this exercise, we calculate the net force on charge \( q_3 \). Since the problem specifies a magnitude for the net force, we use:\[4.00 \times 10^{-6} \, \text{N} = \left|F_{31} + F_{32}\right| \]Through this relation, we substitute and calculate the expressions for \( F_{31} \) and \( F_{32} \) derived in previous steps.
  • Calculate \( F_{31} \) using position and magnitude of \( q_1 \) relative to \( q_3 \)
  • Calculate \( F_{32} \) using position and magnitude of \( q_2 \) relative to \( q_3 \)
By doing this, we effectively evaluate the scenario to ensure the net force experienced by \( q_3 \) is set appropriately and different responses can be made on its potential value and placement.
Force Equilibrium
Force equilibrium occurs when the sum of all forces acting on an object equals zero. This means the object is in a steady state without acceleration.

In part (c) of our problem, the focus is finding where charge \( q_3 \) can be placed such that it achieves force equilibrium. The condition for equilibrium in this case is:\[F_{31} = F_{32}\]This implies the force exerted by charge \( q_1 \) on charge \( q_3 \) equals the force exerted by charge \( q_2 \) on charge \( q_3 \). To solve this:
  • Set the expressions for \( F_{31} \) and \( F_{32} \) equal.
  • Utilize the specific positions of \( q_1 \) and \( q_2 \) to derive the relationship.
  • Solve the resulting equation for \( x \), identifying the necessary position for \( q_3 \).
This process ensures understanding of how electric forces balance each other to bring about a state of no net force on the charge.

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Most popular questions from this chapter

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(r\) in the direction of (a) the field point at \(x=0, \quad y=-1.35 \mathrm{m}\) ; (b) the field point at \(x=12.0 \mathrm{cm}, y=12.0 \mathrm{cm} ;(\mathrm{c})\) the field point at \(x=-1.10 \mathrm{m},\) \(y=2.60 \mathrm{m} ?\) Express your results in terms of the unit vectors \(\hat{\boldsymbol{\imath}}\) and \(\hat{\boldsymbol{J.}}\)

A negative point charge \(q_{1}=-4.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=0.60 \mathrm{m} .\) A second point charge \(q_{2}\) is on the \(x\) -axis at \(x=-1.20 \mathrm{m} .\) What must the sign and magnitude of \(q_{2}\) be for the net electric field at the origin to be (a) 50.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -direction and \((\mathrm{b}) 50.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) -x-direction?

CP A thin disk with a circular hole at its center, called an annulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. P21. 104 ). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z-\) plane, with its center at the origin. For an arbitrary point on the \(x\) -axis (the axis of the annulus), find the magnitude and direction of the electric field \(\vec{E} .\) Consider points both above and below the annulus in Fig. P21. \(104 .\) (c) Show that at points on the \(x\) -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\) -axis (but cannot move off the axis. The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section \(14.2 .\) The annulus is held stationary.)

CP A small sphere with mass \(m\) carries a positive charge \(q\) and is attached to one end of a silk fiber of length \(L .\) The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density \(\sigma\) . Show that when the sphere is in equilibrium, the fiber makes an angle equal to arctan \(\left(q \sigma / 2 m g \epsilon_{0}\right)\) with the vertical sheet.

Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. The maximum rate of charge flow in a lightning bolt is about \(20,000 \mathrm{C} / \mathrm{s} ;\) this lasts for 100\(\mu\) or less. How much charge flows between the ground and the cloud in this time? How many electrons flow during this time?
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