Problem 76 - University Physics with Modern Physics

Question

Two identical spheres are each attached to silk threads of length \(L=0.500\) in and hung from a common point (Fig. 21.44\()\) . Each sphere has mass \(m=8.00 \mathrm{g}\) . The radius of each sphere isvery small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_{1}\) , and the other a different positive charge \(q_{2}\) ; this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta=20.0^{\circ}\) with the vertical. (a) Draw afree-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the information you have been given, what can you say about the magnitudes of \(q_{1}\) and \(q_{2} ?\)Explain your answers. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of \(30.0^{\circ}\) with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

Step-by-Step Solution

Answer

The electrostatic force is \(F_e = 0.0294 \text{ N}\); tension is \(T = 0.0832 \text{ N}\). The original charges satisfy \(q_1 q_2 > 0\).

Step 1: Analyze the forces acting on the spheres

Begin by drawing a free-body diagram for each sphere. The forces acting on each sphere are the tension in the thread \(T\), the gravitational force \(mg\), and the electrostatic force \(F_e\). The tension \(T\) can be divided into its horizontal component \(T \sin \theta\) and vertical component \(T \cos \theta\). The electrostatic force and horizontal component of the tension are equal, as are the gravitational force and vertical component of the tension.

Step 2: Establish equilibrium equations

Since the spheres are in equilibrium, the net force in both the horizontal and vertical directions must be zero. We can write:- For vertical equilibrium: \( T \cos \theta = mg \) - For horizontal equilibrium: \( T \sin \theta = F_e \)

Step 3: Solve for the electrostatic force

From the vertical equilibrium \( T \cos \theta = mg \), solve for \(T\): \[ T = \frac{mg}{\cos \theta} \]Now use horizontal equilibrium equation \( T \sin \theta = F_e \) to find \(F_e\): \[ F_e = T \sin \theta = \frac{mg \sin \theta}{\cos \theta} = mg \tan \theta \]Substitute \(m = 0.008 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(\theta = 20.0^\circ\) to find \(F_e\).

Step 4: Calculate the tension in each thread

Substitute the known values into \( T = \frac{mg}{\cos \theta} \) to find the tension. We already have that \(m = 0.008 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(\theta = 20.0^\circ\). Calculate \(T\).

Step 5: Discussion about the charges

Since the spheres are initially separated and exert equal forces on each other, \(q_1\) and \(q_2\) should satisfy the condition \(q_1 q_2 > 0\). However, their magnitudes may be different since only equilibrium data is given.

Step 6: Use conservation of charge for equal charges after wire connection

When a wire is connected and then removed, each sphere has charge \(q\). The new angle with the vertical is \(30.0^\circ\). The principle of charge conservation gives: \[q_1 + q_2 = 2q\]Use the new forces situation: \[ F_e = mg \tan 30.0^\circ \] Integrate Coulomb's law and solve for the individual charges.

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics that describes how charged objects interact. According to this law, the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, this can be represented as:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:

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

In our exercise, each sphere is treated as a point charge, and we use Coulomb's Law to understand the electrostatic force acting on each sphere, which is crucial for determining the forces when the spheres reach equilibrium.

Equilibrium Forces

When objects are in equilibrium, they experience no net force. This means all the forces acting on them are balanced. In the context of this exercise, the spheres are in equilibrium when suspended, indicating that the forces in both the vertical and horizontal directions are equal and opposite.
In the vertical direction, the tension in the thread created by the spheres ( \( T \cos \theta \) ) counterbalances the gravitational force ( \( mg \) ). Hence:\[ T \cos \theta = mg \]In the horizontal direction, the electrostatic force ( \( F_e \) ) is balanced by the horizontal component of the tension force ( \( T \sin \theta \) ). Therefore:\[ T \sin \theta = F_e \]
Balance in these forces is essential for the spheres to maintain their positions when hanging from the threads.

Charge Conservation

The principle of charge conservation states that the total electric charge in an isolated system remains constant. In this exercise, when a wire connects the two spheres, allowing them to share charge, the total charge before and after this connection remains the same.
Initially, the charges are \( q_1 \) and \( q_2 \) and after connection, both spheres have equal charges ( \( q \) ). This can be written as:\[ q_1 + q_2 = 2q \]
By utilizing the charge conservation principle, you can determine the individual charges before total transfer. This conservation provides clues to solve complex systems where charges move and redistribute among components.

Free-body Diagram

Creating a free-body diagram is an essential step when analyzing forces acting on an object. This diagram provides a clear visual representation of all the forces involved, which aids in solving for unknowns like tension or electrostatic force.
For each sphere in the exercise, draw:

• Gravitational force ( \( mg \) ) acting vertically down.
• Tension ( \( T \) ) in the thread, acting through the string and having both vertical ( \( T \cos \theta \) ) and horizontal components ( \( T \sin \theta \) ).
• Electrostatic force ( \( F_e \) ) due to other sphere, acting horizontally.

The balancing of these forces, shown in the free-body diagram, helps set the stage for writing equilibrium equations.