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Chapter 21: Problem 17

Three point charges are arranged along the \(x\)-axis. Charge \(q_1 = +3.00 \space \mu\)C is at the origin, and charge \(q_2 = -5.00 \space \mu\)C is at \(x =\) 0.200 m. Charge \(q_3 = -8.00 \space \mu\)C. Where is \(q_3\) located if the net force on \(q_1\) is 7.00 N in the \(-\) \(x\)-direction ?

Short Answer

Expert verified
Charge \(q_3\) is located at \( x = -0.456 \text{ m} \) along the \(x\)-axis.

Step by step solution

01

Understand the Forces Involved

We have three point charges, with their interactions governed by Coulomb's Law, which states that the electric force between two charges is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The net force on charge \(q_1\) is given by the forces exerted by \(q_2\) and \(q_3\) on \(q_1\).
02

Coulomb's Law Equation

The force exerted by a charge \(q\) on another charge \(Q\), separated by a distance \(r\), is given by: \[ F = k \frac{|qQ|}{r^2} \]where \(k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\). This will be used to calculate individual forces \(F_{12}\) between \(q_1\) and \(q_2\), and \(F_{13}\) between \(q_1\) and \(q_3\).
03

Calculate Force due to q2 on q1 ( F_{12})

Using Coulomb's Law for \(F_{12}\):\[ F_{12} = k \frac{|q_1 \times q_2|}{(0.200)^2} \]Substitute the values:\[ F_{12} = (8.99 \times 10^9) \frac{3.00 \times 10^{-6} \times 5.00 \times 10^{-6}}{0.200^2} \approx 3.37 \, \text{N} \] This force is in the positive \(x\)-direction, since they have opposite signs.
04

Determine the Force due to q3 on q1 (F_{13})

Since the net force on \(q_1\) is 7.00 N in the \(-x\)-direction, we can express the total force equation:\[ F_{net} = F_{12} - F_{13} = -7.00 \, \text{N} \]Re-arrange this to find \(F_{13}\):\[ F_{13} = F_{12} + 7.00 = 3.37 + 7.00 = 10.37 \, \text{N} \]
05

Find the Distance of q3 from q1

To locate \(q_3\), use Coulomb's Law:\[ F_{13} = k \frac{|q_1 \times q_3|}{x^2} \]Substitute the known values:\[ 10.37 = (8.99 \times 10^9) \frac{3.00 \times 10^{-6} \times 8.00 \times 10^{-6}}{x^2} \]Solving for \(x\), we get:\[ x^2 = \frac{(8.99 \times 10^9) \times 3.00 \times 10^{-6} \times 8.00 \times 10^{-6}}{10.37} \]\[ x^2 = 20.756 \times 10^{-2} \]\[ x \approx 0.456 \text{ m} \]Since this force is in the negative \(x\)-direction, \(q_3\) is placed at \( x = -0.456 \text{ m} \).

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Key Concepts

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

Electric Force
The concept of electric force is fundamentally linked to the interactions between charged particles. In physics, the electric force that acts between two point charges is governed by Coulomb's Law. This foundational principle states that the force is directly proportional to the product of the magnitudes of the charges involved and inversely proportional to the square of the distance separating them.

In more straightforward terms, the larger the charge magnitudes or the closer they are, the stronger the force. Electric force can be attractive or repulsive. It becomes attractive when charges have opposite signs and repulsive when charges have the same signs.

Electric forces play a key role in many natural phenomena and technological applications, from the operation of electronic devices to chemical bonding and beyond.
Point Charges
Point charges are idealized charges that are considered to be located at a single point in space. These charges simplify the calculations in electrostatics because their small size allows us to ignore the complexities of charge distribution on objects.

Point charges are often used in theoretical models to help us understand the behavior of electric fields and forces in a simplified manner. For example, in the original exercise, the charges are treated as point charges, so we can easily calculate the forces using Coulomb's Law.

The notion of point charges is crucial in electrostatics as it provides the basis for determining how charges interact in various configurations, such as the alignment along a line or in a plane.
Net Force Calculation
Net force is a critical concept that refers to the total force acting on an object when all the individual forces are considered. It determines how an object will move or stand still.

When multiple forces act on a charge, you need to consider all contributions to find the net force. This can involve summing the vector forces because directionality is crucial. For instance, a positive force in one direction will offset a negative one in the opposite direction.

In the exercise at hand, calculating the net force was necessary to determine the position of one of the charges. By analyzing the forces from other charges and applying the net force equation, it became possible to deduce unknown distances or positions.
Electric Charge Interaction
Electric charge interactions are fundamental to understanding how charges exert forces on each other within a system. Different charges interact based on their polarity: positive with negative charges will attract, whereas like charges will repel.

This interaction is quantifiable by Coulomb's Law, enabling predictions of force magnitudes and directions. Moreover, such interactions underpin the behavior of atoms, molecules, and electromagnetic fields.

In practical terms, comprehending these interactions helps in predicting the stability of chemical compounds, the alignment of electronic circuits, and even celestial phenomena. In our specific case of point charges aligned on the x-axis, their interactions determine how they exert forces on one another, guiding us to locate where one charge must be positioned.

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Most popular questions from this chapter

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

A proton is placed in a uniform electric field of 2.75 \(\times 10^3 \space N/C\). Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00 \(\mu\)s in the field, assuming it starts from rest.

Suppose you had two small boxes, each containing 1.0 g of protons. (a) If one were placed on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? Express your answer in newtons and in pounds. Do you need to take into account the gravitational forces of the earth and moon on the protons? Why? (b) What gravitational force would each box of protons exert on the other box?

A negative point charge \(q_1 = -4.00\) nC is on the \(x\)-axis at \(x =\) 0.60 m. A second point charge \(q_2\) is on the \(x\)-axis at \(x = -\)1.20 m. What must the sign and magnitude of \(q_2\) be for the net electric field at the origin to be (a) 50.0 N\(/\)C in the \(+x\)-direction and (b) 50.0 N\(/\)C in the \(-\)x-direction?

A nerve signal is transmitted through a neuron when an excess of \(Na^+\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately 10.0 \(\mu\)m in diameter, and measurements show that about 5.6 \(\times \space 10^{11} \space Na^+\)ions per meter (each of charge \(+e\)) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the axon. Consider a 0.10-mm length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.10-mm length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 cm below the skin? (c) Certain sharks can respond to electric fields as weak as 1.0 \(\mu N/C\). How far from this segment of axon could a shark be and still detect its electric field?
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