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Chapter 23: Problem 51

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?

Short Answer

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

Step by step solution

01

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.
02

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 \( x = 10.0 \text{ cm} \).
03

Plug in Values for Part (a)

Using the expression from Step 2, calculate the potential energy when \( q_3 \) is at 10.0 cm: \[ U = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \left( \frac{(4.00 \times 10^{-9} \text{ C})(-3.00 \times 10^{-9} \text{ C})}{0.20} + \frac{(4.00 \times 10^{-9} \text{ C})(2.00 \times 10^{-9} \text{ C})}{0.10} + \frac{(-3.00 \times 10^{-9} \text{ C})(2.00 \times 10^{-9} \text{ C})}{0.10} \right) \]
04

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} \.
05

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 \( r_{23} = 20 - x_3 \).
06

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 \( x_3 = 12 \text{ cm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) Show that \(V\) for a spherical shell of radius \(R,\) that has charge \(q\) distributed uniformly over its surface, is the same as \(V\) for a solid conductor with radius \(R\) and charge \(q .\) (b) You rub an inflated balloon on the carpet and it acquires a potential that is 1560 \(\mathrm{V}\) lower than its potential before it became charged. If the charge is uniformly distributed over the surface of the balloon and if the radius of the balloon is \(15 \mathrm{cm},\) what is the net charge on the balloon? (c) In light of its \(1200-\mathrm{V}\) potential difference relative to you, do you think this balloon is dangerous? Explain.

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{cm} . \mathrm{A}\) small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 \(\mathrm{cm}\) of the surface of the large sphere?

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 \(\mathrm{cm}\) in diameter, has mass \(50.0 \mathrm{g},\) and contains \(-10.0 \mu \mathrm{C}\) of charge. The other sphere is 40.0 \(\mathrm{cm}\) in diameter, has mass \(150.0 \mathrm{g},\) and contains \(-30.0 \mu \mathrm{c}\) of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (Hint: The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)

Self-Energy of a Sphere of Charge. A solid sphere of radius \(R\) contains a total charge \(Q\) distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the "self-energy" of the charge distribution. (Hint: After you have assembled a charge \(q\) in a sphere of radius \(r,\) how much energy would it take to add a spherical shell of thickness dr having charge \(d q ?\) Then integrate to get the total energy.)

A total electric charge of 3.50 \(\mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of 24.0 \(\mathrm{cm}\) . If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{cm} ;\) (b) \(24.0 \mathrm{cm} ;(\mathrm{c}) 12.0 \mathrm{cm}.\)
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