Problem 93 - University Physics with Modern Physics

Question

Two small spheres, each carrying a net positive charge, are separated by \(0.400 m\). You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (\(charge \space q_1\)) at the origin and the other sphere (\(charge \space q_2\)) at \(x = +\)0.400 m. Available to you are a third sphere with net charge \(q_3 = 4.00 \times 10^{-6}\) C and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the \(x\)-axis at \(x =\) 0.200 m; you measure the net force on it to be 4.50 N in the \(+ x\)-direction. Then you move the third sphere to \(x = +\)0.600 m and measure the net force on it now to be 3.50 N in the \(+ x\)-direction. (a) Calculate \(q_1\) and \(q_2\). (b) What is the net force (magnitude and direction) on \(q_3\) if it is placed on the \(x\)-axis at \(x = -\)0.200 m? (c) At what value of \(x\) (other than \(x = \pm \infty\)) could \(q_3\) be placed so that the net force on it is zero?

Step-by-Step Solution

Answer

(a) Solve for \(q_1\) and \(q_2\) from the equations. (b) Calculate \(F_{net}\) using positions. (c) Find \(x\) from equation balancing forces.

Step 1: Analyze First Force Condition

When the third sphere with charge \(q_3\) is placed at \(x = 0.200 \text{ m}\), the net force on it is \(4.50 \text{ N}\) in the \(+x\)-direction. Using Coulomb's Law, the force due to \(q_1\) at the origin is \(F_{31} = k \cdot \frac{|q_3 q_1|}{0.200^2}\) and due to \(q_2\) at \(x = 0.400 \text{ m}\) is \(F_{32} = k \cdot \frac{|q_3 q_2|}{0.200^2}\). The net force \(F_{net} = F_{32} - F_{31} = 4.50 \, \text{N}\).

Step 2: Analyze Second Force Condition

When \(q_3\) is placed at \(x = 0.600 \, \text{m}\), the net force is \(3.50 \, \text{N}\) in the \(+x\)-direction. Again, applying Coulomb's Law, the force due to \(q_1\) is \(F_{31} = k \cdot \frac{|q_3 q_1|}{0.600^2}\) and due to \(q_2\), it is \(F_{32} = k \cdot \frac{|q_3 q_2|}{0.200^2}\). Thus, \(F_{net} = F_{32} + F_{31} = 3.50 \, \text{N}\).

Step 3: Set Up Equations

From Step 1, we have: \[ k \cdot \frac{q_3 q_2}{0.200^2} - k \cdot \frac{q_3 q_1}{0.200^2} = 4.50 \text{ N} \] From Step 2: \[ k \cdot \frac{q_3 q_2}{0.200^2} + k \cdot \frac{q_3 q_1}{0.600^2} = 3.50 \text{ N} \]

Step 4: Simplify and Solve Equations

With \(k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\) and \(q_3 = 4.00 \times 10^{-6} \, \text{C}\), plug these values into the equations:1. \( F_{32} - F_{31} = 4.50 \) gives: \( 179.8 \cdot q_2 - 44.95 \cdot q_1 = 4.50 \)2. \( F_{32} + F_{31} = 3.50 \) gives: \( 179.8 \cdot q_2 + 4.99 \cdot q_1 = 3.50 \)Solve these two equations to find \(q_1\) and \(q_2\).

Step 5: Calculate Net Force on q3 at -0.200 m

Now place \(q_3\) at \(x = -0.200 \,\text{m}\) and calculate net force. The distance to \(q_1\) is \(0.200 \,\text{m}\), and to \(q_2\) is \(0.600 \,\text{m}\). Thus, force due to \(q_1\) using Coulomb's Law is \(F_{31} = k \cdot \frac{|q_3 q_1|}{0.200^2}\) and since \(q_3\) is on the \(-x\)-axis, the force by \(q_2\) is \(F_{32} = k \cdot \frac{|q_3 q_2|}{0.800^2}\). Calculate \(F_{net} = F_{31} - F_{32}\).

Step 6: Find the Equilibrium Position

To find the point where the net force on \(q_3\) is zero, set the positions equal in the force equations such that forces cancel each other out. Let \(x\) be the position on the \(x\)-axis where \(q_3\) experiences zero net force. Set \[ F_{31} = F_{32} \to k \cdot \frac{|q_3 q_1|}{x^2} = k \cdot \frac{|q_3 q_2|}{(0.400-x)^2} \] Solve for \(x\).

Key Concepts

Electric Force

Electric force is a fundamental interaction between charges. When dealing with charges, it's important to remember Coulomb's Law, which states that the electric force \( F \) between two point charges \( q_1 \) and \( q_2 \) is directly proportional to the product of the absolute values of those charges and inversely proportional to the square of the distance \( r \) between them. This can be mathematically represented by the formula: \[ F = k \frac{|q_1 q_2|}{r^2} \]where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2. \) Coulomb's Law helps us understand how charges interact based on their magnitudes and the distances between them. - The electric force is attractive if the charges are opposite.- The force is repulsive if the charges are the same.Since electric forces are vectors, they have both magnitude and direction, which is crucial when determining the net force on a charge when multiple other charges are present. Applying this law involves:- Calculating individual electric forces using the formula.- Adding vectorially to find the net force acting on a charge due to others.
Understanding the nature of electric forces is pivotal in solving physics problems involving charged objects.

Charge Interaction

Charge interaction refers to how charged particles influence each other through electric forces. The behavior of charges can be understood by observing the principles of repulsion and attraction. Opposite charges attract, pulling towards each other, while like charges repel, pushing away from one another. This behavior forms the basis of many electrical phenomena. In scenarios such as the one given in the exercise, it involves analyzing how one or more charged objects will affect a test charge placed within their electric fields.
To understand charge interactions:

• Identify the charges involved and their signs (positive or negative).
• Recognize the distances between each pair of charges.
• Employ Coulomb's Law to compute the forces between the charges.

Charge interaction is not always intuitive because the forces act over a distance and can change with position. When multiple charges are present, their individual electric fields superimpose, affecting the resulting force on any test charge in the vicinity. Thus, real-world applications require combining these interactions to predict how charged bodies will move or stay in equilibrium based on the variables of charge quantity, spatial orientation, and magnitude.

Physics Problem Solving

Solving physics problems involving electric forces and charge interactions often requires a systematic approach. First, understanding the given situation and the relationships between charges is necessary to set up relevant equations accurately. Let's break down how to approach such problems efficiently: 1. **Understand the Problem**: Begin by identifying the known and unknown quantities. This includes the magnitudes of charges, distances between them, and any forces involved. Visualizing the problem through sketches can be immensely helpful. 2. **Apply Relevant Physics Laws**: In this context, use Coulomb's Law to derive expressions for forces between different pairs of charges. This will often involve setting up equations based on the given positions and measured forces. 3. **Solve the Equations**: With the equations ready, use algebra to solve them for the unknown quantities. In our exercise, we set simultaneous equations for different force conditions, which are then solved to find unknown charges. 4. **Verify and Interpret the Result**: Once the solution is obtained, it's crucial to check if it makes sense physically. Are the directions and magnitudes of the forces plausible given the configuration of charges? Physics problem-solving, particularly in the realm of electric interactions, builds critical thinking and logical reasoning skills. Incorporating clear steps:

• Helps break down complex problems into simpler parts.
• Makes it easier to detect errors and make necessary corrections.

Using such an approach facilitates comprehending not only the mathematical side of the problem but also the physical interactions at play.
This holistic understanding is invaluable in solving both academic exercises and real-world scenarios.