Problem 51 - University Physics with Modern Physics

Question

A point charge \(q_{1}=4.00 \mathrm{nC}\) is placed at the origin, and a second point charge \(q_{2}=-3.00 \mathrm{nC}\) is placed on the \(x\) -axis at \(x=+20.0 \mathrm{cm} .\) A third point charge \(q_{3}=2.00 \mathrm{nC}\) is to be placed on the \(x\) -axis between \(q_{1}\) and \(q_{2} .\) (Take as zero the potential energy of the three charges when they are infinitely far apart. (a) What is the potential energy of the system of the three charges if \(q_{3}\) is placed at \(x=+10.0 \mathrm{cm} ?\) (b) Where should \(q_{3}\) be placed to make the potential energy of the system equal to zero?

Step-by-Step Solution

Answer

(a) The potential energy is \(-359.6 \text{ nJ}\). (b) Charge \(q_3\) should be at 12 cm.

Step 1: Understand the Setup

The problem involves three point charges along the x-axis: \( q_1 = 4.00 \text{ nC} \) at the origin, \( q_2 = -3.00 \text{ nC} \) at \( x = 20.0 \text{ cm} \), and \( q_3 = 2.00 \text{ nC} \) which needs to be placed in between \( q_1 \) and \( q_2 \). We need to calculate the potential energy of the system based on their arrangement.

Step 2: Calculate Potential Energy Expression

The potential energy \( U \) for a system of point charges is given by the sum of potential energies due to pairs of charges: \[ U = k_e \left( \frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}} \right) \] where \( k_e \) is Coulomb's constant, and \( r_{ij} \) is the distance between charges \( q_i \) and \( q_j \). Here, \( r_{12} = 0.20 \text{ m} \), \( r_{13} = 0.10 \text{ m} \), \( r_{23} = 0.10 \text{ m} \) when \( q_3 \) is at \( x = 10.0 \text{ cm} \).

Step 3: Plug in Values for Part (a)

Using the expression from Step 2, calculate the potential energy when \( q_3 \) is at 10.0 cm: \[ U = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \left( \frac{(4.00 \times 10^{-9} \text{ C})(-3.00 \times 10^{-9} \text{ C})}{0.20} + \frac{(4.00 \times 10^{-9} \text{ C})(2.00 \times 10^{-9} \text{ C})}{0.10} + \frac{(-3.00 \times 10^{-9} \text{ C})(2.00 \times 10^{-9} \text{ C})}{0.10} \right) \]

Step 4: Simplify Expression for Part (a)

Calculate each term: \( \frac{4.00 \times (-3.00)}{0.20} = -60 \), \( \frac{4.00 \times 2.00}{0.10} = 80 \), and \( \frac{-3.00 \times 2.00}{0.10} = -60 \). Hence, \( U = 8.99 \times 10^9 \left( -60 + 80 - 60 \right) \)] = 8.99 \times 10^9 \times (-40) = -359.6 \text{ nJ} \.

Step 5: Solve for Zero Potential Energy in Part (b)

We want the potential energy \( U \) to be zero. Set up the equation: \[ \frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}} = 0 \] with known \( q_i \). Assume \( q_3 \) is placed at \( x = x_3 \text{ cm} \). This gives \( r_{13} = x_3 \) and \( r_{23} = 20 - x_3 \).

Step 6: Simplify and solve for \( x_3 \)

Plug into zero-energy condition: \[ \frac{(4.00)(-3.00)}{0.20} + \frac{4.00 \times 2.00}{x_3} + \frac{-3.00 \times 2.00}{20.0 - x_3} = 0 \]. Solving gives \( \frac{8}{x_3} + \frac{-6}{20 - x_3} = 60 \). This can be solved to find \( x_3 = 12 \text{ cm} \).

Key Concepts

Point Charge

A point charge is a fundamental concept in electrostatics where an electric charge is considered to be concentrated at a single point in space. While this is an idealization, it helps to simplify many problems in electrostatics. By treating charges as point-like, we can analyze the interactions between multiple charges without worrying about their physical size or shape.

• Characteristics: Point charges are assumed to have no dimensions and are solely defined by their magnitude and sign.
• Uses: They are often used in physics to model spherical charge distributions, where the sphere is small enough compared to the distance between interacting charges.
• In Application: In the problem, three point charges are involved: one at the origin, another 20 cm along the x-axis, and a third positioned between the two.

Understanding point charges is critical as they form the basis for defining electric fields, forces between charges, and potential energy of a charge configuration.
They help us simplify calculations and understand the fundamental concepts underlying electrostatic interactions.

Potential Energy

In electrostatics, potential energy refers to the energy stored within a system of charges due to their relative positions. It is a scalar quantity that depends on the configuration of the charges and the distances between them. The concept of potential energy is crucial for understanding how energy is stored and transferred in electric fields.

• Equation: For a system of point charges, the potential energy is calculated by the formula: \[ U = k_e \left( \frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}} \right) \] where \( q_i \) are the charges, \( r_{ij} \) are the distances between the charges, and \( k_e \) is Coulomb's constant.
• Importance: This concept helps identify the energy dynamics within an electric field and is essential for understanding how charges interact and affect each other's motion.
• Zero Reference Point: The potential energy is often considered to be zero when charges are infinitely far apart. This simplifies calculations, allowing us to focus solely on the energy changes due to their positions.

In the given problem, potential energy helps determine where to place the third charge such that the overall energy of the system is zero. This requires understanding how energy contributions from each pair of charges add up or negate each other.

Coulomb's Law

Coulomb's Law is a fundamental principle that describes the force between two point charges. It provides the basis for calculating the electric force and thereby interactions in electrostatics. The law states that the force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

• Mathematical Expression: \( F = k_e \frac{q_1q_2}{r^2} \)where \( F \) is the force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between them, and \( k_e \) is Coulomb's constant.
• Vector Nature: The force is a vector, meaning it has both magnitude and direction. The direction is along the line joining the charges, and the force is attractive if the charges have opposite signs and repulsive if they are the same sign.
• Applications: In problems like the one described, Coulomb's law allows us to calculate not just forces, but also to extend to potential energies by integrating the force over distance.

Understanding Coulomb's Law is essential for solving problems involving electric charges and predicting how they will interact within a field. It is the foundation for further exploration of electric fields and potential energy, which are key in more complex electrostatic scenarios.