Problem 19 - University Physics with Modern Physics

Question

Three point charges are arranged along the \(x\) -axis. Charge \(q_{1}=+3.00 \mu C\) is at the origin, and charge \(q_{2}=-5.00 \mu C\) is at \(x=0.200 \mathrm{m} .\) Charge \(q_{3}=-8.00 \mu \mathrm{C} .\) Where is \(q_{3}\) located if the net force on \(q_{1}\) is 7.00 \(\mathrm{N}\) in the \(-x\) -direction?

Step-by-Step Solution

Answer

Charge \( q_3 \) is located at \(-0.159 \, \mathrm{m}\) on the x-axis.

Step 1: Understand the Problem

We are given three charges along the x-axis. Charge \( q_1 = +3.00\, \mu C \) is at the origin, charge \( q_2 = -5.00\, \mu C \) is located at \( x = 0.200 \,\mathrm{m} \), and charge \( q_3 = -8.00\, \mu C \) needs to be located such that the net force on \( q_1 \) is \( 7.00 \, \mathrm{N} \) in the \(-x\)-direction.

Step 2: Use Coulomb's Law

The force between two charges \( q_i \) and \( q_j \) separated by a distance \( r \) is given by Coulomb's law: \[ F = k \frac{|q_i q_j|}{r^2} \]where \( k = 8.99 \times 10^9 \, \mathrm{N\cdot m^2/C^2} \) is Coulomb's constant. Apply this to find forces on \( q_1 \) due to \( q_2 \) and \( q_3 \).

Step 3: Calculate Force by q2 on q1

The force exerted by \( q_2 \) on \( q_1 \) is:\[ F_{12} = k \frac{|q_1 \cdot q_2|}{(0.200)^2} \]Calculate:\[ F_{12} = 8.99 \times 10^9 \frac{|3.00 \times 10^{-6} \cdot (-5.00 \times 10^{-6})|}{0.200^2} \approx 3.37 \, \mathrm{N} \] in the \(+x\)-direction.

Step 4: Set Up Net Force Equation

Since the net force on \( q_1 \) is \( 7.00 \, \mathrm{N} \) in the \(-x\)-direction:\[ F_{13} - F_{12} = 7.00 \, \mathrm{N} \]where \( F_{13} \) is the force exerted by \( q_3 \) on \( q_1 \).

Step 5: Solve for Force F_{13}

\( F_{13} \) has to be enough to counteract \( F_{12} \) and contribute to the net force:\[ F_{13} = 7.00 \, \mathrm{N} + 3.37 \, \mathrm{N} = 10.37 \, \mathrm{N} \]The force thus has to be \(10.37 \, \mathrm{N}\) in the \(-x\)-direction.

Step 6: Calculate Position of q3

Use Coulomb's law to find the position \( x \) of \( q_3 \):\[ 10.37 = 8.99 \times 10^9 \frac{|3.00 \times 10^{-6} \times 8.00 \times 10^{-6}|}{x^2} \]Solve for \( x \):\[ x^2 = 8.99 \times 10^9 \frac{24.00 \times 10^{-12}}{10.37} \]Calculating gives \( x \approx 0.159 \, \mathrm{m} \). Since the net force is in the \(-x\) direction, \( q_3 \) must be positioned at \( x = -0.159 \, \mathrm{m} \).

Key Concepts

Electric Force

The electric force is a fundamental interaction between charged objects caused by their electric charges. It is responsible for binding atoms, molecules, and even macroscopic materials together. The electric force can be either attractive or repulsive, depending on the nature of the charges involved.

When dealing with point charges, you can predict the force using Coulomb's Law, which is expressed as: \[ F = k \frac{|q_i q_j|}{r^2} \] where:

• \( F \) is the magnitude of the electric force between two charges.
• \( k = 8.99 \times 10^9 \, \mathrm{N\cdot m^2/C^2} \) is Coulomb's constant, a value that represents the strength of the electric force in a vacuum.
• \( q_i \) and \( q_j \) are the magnitudes of the charges in Coulombs.
• \( r \) is the distance between the charges.

Coulomb's Law showcases that the electric force is inversely proportional to the square of the distance between the charges. This means, as the distance increases, the force rapidly weakens.

In the given problem, you must analyze the forces exerted on a specific charge due to other charges in its vicinity. Mastery of electric force concepts allows predicting the behavior of charged systems, crucial for understanding electric fields and potential.

Point Charges

Point charges are idealized models of charged objects where the size of the charge is considered to be infinitesimally small compared to the distances involved in the problem. This simplification allows the implementation of Coulomb's Law in calculations without worrying about the shape or volume of the charges.

In our exercise, we consider three point charges arranged along a line on the x-axis:

• \( q_1 = +3.00 \, \mu C \), which is positioned at the origin.
• \( q_2 = -5.00 \, \mu C \), located at \( x = 0.200 \, \mathrm{m} \).
• \( q_3 = -8.00 \, \mu C \), whose location we need to determine.

The use of point charges is especially useful in theoretical scenarios and simplifications where the details of the charge distribution are not significant for calculations.

Understanding point charges helps us focus simply on how charged objects interact at distances, making complex charge interactions solvable through a series of simple point interactions. This forms the backbone of electrostatics and helps in designing electronic components and understanding phenomena at atomic and molecular scales.

Net Force

The concept of net force is about understanding the overall force acting on an object when multiple forces are applied. It involves vector addition because force is a vector quantity, meaning it has both magnitude and direction.

For our problem, point charge \( q_1 \) is influenced by forces from both \( q_2 \) and \( q_3 \). The net force on \( q_1 \) was given as \( 7.00 \, \mathrm{N} \) in the \(-x\)-direction. Knowing the force from \( q_2 \) on \( q_1 \) allowed us to calculate what the force from \( q_3 \) must be.

To find the net force:

• Calculate the individual forces on the charge in question.
• Combine these forces vectorially, considering their directions – in this case, the directions are along the x-axis.

The equation for the net force in this context is: \[ F_{13} - F_{12} = 7.00 \, \mathrm{N} \] indicating that \( F_{13} \), the force from \( q_3 \) on \( q_1 \), must be stronger than \( F_{12} \), the force from \( q_2 \) on \( q_1 \), to result in a net force in the \(-x\)-direction.

Understanding net force and its calculation is crucial in problems involving dynamics and helps predict how an object will move or stay in equilibrium under multiple forces.