Problem 68 - College Physics

Question

\(\bullet$$\bullet\) A charge of \(-3.00 \mathrm{nC}\) is placed at the origin of an \(x y-\)coordi- nate system, and a charge of 2.00 \(\mathrm{nC}\) is placed on the \(y\) axis at \(y=4.00 \mathrm{cm} .\) (a) If a third charge, of \(5.00 \mathrm{nC},\) is now placed at the point \(x=3.00 \mathrm{cm}, y=4.00 \mathrm{cm},\) find the \(x\) and \(y\) com- ponents of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

Step-by-Step Solution

Answer

Net force is \(4.60 \times 10^{-5} \, \text{N}\), rightward along the x-axis.

Step 1: Understand the positions of charges

The first charge, \( q_1 = -3.00 \, \text{nC} \), is at the origin (0, 0), the second charge, \( q_2 = 2.00 \, \text{nC} \), is at \( (0, 4.00) \, \text{cm} \). The third charge, \( q_3 = 5.00 \, \text{nC} \), is placed at \( (3.00, 4.00) \, \text{cm} \).

Step 2: Calculate the distance between charges

Compute the distance between \( q_1 \) and \( q_3 \) using the formula \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):For \( q_1 \) and \( q_3 \): \\[ r_{13} = \sqrt{(3.00)^2 + (4.00)^2} = 5.00 \, \text{cm} \]For \( q_2 \) and \( q_3 \): \\[ r_{23} = \sqrt{(3.00 - 0.00)^2 + (4.00 - 4.00)^2} = 3.00 \, \text{cm} \]

Step 3: Calculate force between q1 and q3 (F13)

Use Coulomb's law: \( F = \frac{k \cdot |q_1 \cdot q_3|}{r_{13}^2} \), where \( k = 8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2 \).\[ F_{13} = \frac{8.99 \times 10^9 \cdot |-3.00 \times 10^{-9} \cdot 5.00 \times 10^{-9}|}{(0.05)^2} = 5.39 \times 10^{-5} \, \text{N} \]

Step 4: Determine the direction of F13

Since both charges are on the same x-axis, \( F_{13} \) acts leftwards (negative x-direction) on \( q_3 \). Thus the direction cosine is used to extract components. \( F_{13, x} = -5.39 \times 10^{-5} \, \text{N} \) and \( F_{13, y} = 0 \).

Step 5: Calculate force between q2 and q3 (F23)

Using Coulomb’s law again:\[ F_{23} = \frac{8.99 \times 10^9 \cdot |2.00 \times 10^{-9} \cdot 5.00 \times 10^{-9}|}{(0.03)^2} = 9.99 \times 10^{-5} \, \text{N} \]

Step 6: Determine direction components of F23

The distance is purely horizontal, so \( F_{23, x} = 9.99 \times 10^{-5} \, \text{N} \), \( F_{23, y} = 0 \).

Step 7: Calculate net force components on q3

Add the vector components: \[ F_{net, x} = F_{13, x} + F_{23, x} = -5.39 \times 10^{-5} + 9.99 \times 10^{-5} = 4.60 \times 10^{-5} \, \text{N} \]\[ F_{net, y} = F_{13, y} + F_{23, y} = 0 + 0 = 0 \]

Step 8: Calculate magnitude of total force

Since the resulting force is purely horizontal, the magnitude equals the x-component:\[ F_{net} = 4.60 \times 10^{-5} \, \text{N} \]

Step 9: Determine the direction

The x-component direction is positive, indicating rightward direction along the x-axis, with angle \( 0^\circ \) from the positive x-axis.

Key Concepts

Electric Force

The electric force is a fundamental interaction between charged particles, described by Coulomb's Law. This force can be attractive or repulsive, depending on the nature of the charges involved. For example, opposite charges attract, while like charges repel each other. Coulomb's Law is formulated as \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] where:

• \(F\) is the magnitude of the force between the charges.
• \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2\).
• \(q_1\) and \(q_2\) are the magnitudes of the charges.
• \(r\) is the distance between the charges.

Understanding this formula is crucial in solving problems involving multiple charges, such as finding the net force on a charge due to the presence of other charges.

Vector Components

In physics, forces are vector quantities, meaning they have both magnitude and direction. To properly analyze forces, especially in two dimensions, it's important to break them into components along defined axes, typically the x and y axes. Vector components allow us to calculate the effect of each force on these axes independently.For instance:

• The x-component \(F_x\) can be calculated using \(F \cdot \cos(\theta)\).
• The y-component \(F_y\) is found with \(F \cdot \sin(\theta)\).

In exercises involving multiple charges, determining the x and y components of electric forces helps in finding the net force on a charge by summing these components. This is essential for both parts (a) and (b) of the original exercise.

Charge Interaction

Charge interaction refers to how different charges affect each other via electric forces. Depending on whether the charges are positive or negative, they will attract or repel one another. The magnitude of the force is determined by both the amounts of charges and how far apart they are.

• Positive-negative charges result in attraction.
• Positive-positive or negative-negative charges result in repulsion.

In the given exercise, the interaction between differently signed charges results in forces with particular directions. Calculating these forces requires understanding the nature of charge interaction and using Coulomb's law to determine the magnitude and direction of forces, which are then broken into components.

Coordinate System

A coordinate system helps define the positions of charges in space, essential for calculating distances and directions when analyzing forces. Typically, a Cartesian coordinate system (x-y plane) is used, where every point is defined by an x and y coordinate. In the problem at hand, charges are positioned relative to this system. The origin is a common reference point. This positioning allows for easy calculation of distances between charges using the distance formula \[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. Using a coordinate system simplifies the complex nature of multiple charge interactions, making it possible to visualize forces and their components effectively. Understanding how to use these systems is vital in physics to solve problems involving forces and motion.