Concept explainers
Review. A block having mass m and charge + Q is connected to an insulating spring having a force constant k. The block lies on a frictionless, insulating, horizontal track, and the system is immersed in a uniform electric field of magnitude E directed as shown in Figure P24.6. The block is released from rest when the spring is unstretched (at x = 0). We wish to show that the ensuing motion of the block is simple harmonic. (a) Consider the system of the block, the spring, and the electric field. Is this system isolated or nonisolated? (b) What kinds of potential energy exist within this system? (c) Call the initial configuration of the system that existing just as the block is released from rest. The final configuration is when the block momentarily comes to rest again. What is the value of x when the block comes to rest momentarily? (d) At some value of x we will call x = x0, the block has zero net force on it. What analysis model describes the particle in this situation? (c) What is the value of x0? (f) Define a new coordinate system x′ such that x′ = x − x0. Show that x′ satisfies a differential equation for
Figure P24.6
(a)
Whether the system is isolated or non-isolated.
Answer to Problem 6P
The system is non-isolated.
Explanation of Solution
Given info: The mass of the block is
The system of the block with a spring is isolated system but when the system of the block with a spring is placed in a electric or magnetic field then it experiences an external force that makes the system non-isolated.
Conclusion:
Therefore, the system is non-isolated.
(b)
The kinds of potential energy exist in the system.
Answer to Problem 6P
The kinds of potential energy exist in the system are elastic potential energy and electrostatic potential energy.
Explanation of Solution
Given info: The mass of the block is
There are two kinds of potential energy exist in the system one is elastic potential energy and other is electrostatic potential energy.
The formula to calculate the elastic potential energy is,
Here,
The formula to calculate the electrostatic potential energy is,
Here,
The formula to calculate electrostatic force is,
Here,
Substitute
Conclusion:
Therefore, the kinds of potential energy exist in the system are elastic potential energy and electrostatic potential energy.
(c)
The value of
Answer to Problem 6P
The value of
Explanation of Solution
Given info: The mass of the block is
The formula to calculate the elastic potential energy is,
Here,
The formula to calculate the electrostatic potential energy is,
Here,
The formula to calculate electrostatic force is,
Here,
Substitute
Equate the formula (1) and (2) as,
Conclusion:
Therefore, the value of
(d)
The analysis model that describes particle in this situation.
Answer to Problem 6P
The analysis model that describes the particle in this situation is in equilibrium.
Explanation of Solution
Given info: The mass of the block is
The above statement is explained as the net force acting on the block is zero at some value of
Conclusion:
Therefore, the particle in this situation is in equilibrium.
(e)
The value of
Answer to Problem 6P
The value of
Explanation of Solution
Given info: The mass of the block is
The formula to calculate the spring force is,
Here,
The formula to calculate electric force is,
Here,
The net force is zero so it is expressed as,
Here,
Rearrange the above expression to find
Conclusion:
Therefore, the value of
(f)
To show: The coordinate
Answer to Problem 6P
Thus, the coordinate
Explanation of Solution
The mass of the block is
Given info:
From part (e), the value of
The formula to calculate the force equation is,
Here,
The formula to calculate the acceleration is,
Here,
The summation of the force is,
Here,
The new coordinate is,
Substitute
Substitute
This is the required equation of S.H.M.
Conclusion:
Therefore, the coordinate
(g)
The period of the simple harmonic motion.
Answer to Problem 6P
The period of the simple harmonic motion is
Explanation of Solution
Given info: The mass of the block is
From part (f), the equation for the S.H.M is,
The general equation of S.H.M is,
Here,
Compare the equation (4) and (5) as,
The formula to calculate the time period is,
Here,
Substitute
Conclusion:
Therefore, the period of the simple harmonic motion is
(h)
The time period depend on the electric field magnitude or not.
Answer to Problem 6P
The time period does not depend on the electric field magnitude.
Explanation of Solution
Given info: The mass of the block is
From part (g),the time period is,
In the above formula, there is no term that signifies electric field magnitude. So the time period is independent of electric field magnitude.
Conclusion:
Therefore, the time period does not depend on the electric field magnitude.
Want to see more full solutions like this?
Chapter 24 Solutions
Physics for Scientists and Engineers with Modern Physics
Additional Science Textbook Solutions
Physical Universe
Conceptual Integrated Science
MODERN PHYSICS (LOOSELEAF)
Essential University Physics: Volume 1 (3rd Edition)
Applied Physics (11th Edition)
Lecture- Tutorials for Introductory Astronomy
- A uniformly charged conducting rod of length = 30.0 cm and charge per unit length = 3.00 105 C/m is placed horizontally at the origin (Fig. P24.37). What is the electric field at point A with coordinates (0, 0.400 m)?arrow_forwardA very long, thin wire fixed along the x axis has a linear charge density of 3.2 C/m. a. Determine the electric field at point P a distance of 0.50 m from the wire. b. If there is a test charge q0 = 12.0 C at point P, what is the magnitude of the net force on this charge? In which direction will the test charge accelerate?arrow_forwardOne end of a light spring with force constant k = 125 N/m is attached to a wall, and the other end to a metal block with charge qA = 2.00 C on a horizontal, frictionless table (Fig. P23.34). A second block with charge qB = 3.60 C is brought close to the first block. The spring stretches as the blocks attract each other so that at equilibrium, the blocks are separated by a distance d = 12.0 cm. What is the displacement x of the spring? Figure P23.34arrow_forward
- Find an expression for the magnitude of the electric field at point A mid-way between the two rings of radius R shown in Figure P24.30. The ring on the left has a uniform charge q1 and the ring on the right has a uniform charge q2. The rings are separated by distance d. Assume the positive x axis points to the right, through the center of the rings. FIGURE P24.30 Problems 30 and 31.arrow_forwardA positively charged disk of radius R = 0.0366 m and total charge 56.8 C lies in the xz plane, centered on the y axis (Fig. P24.35). Also centered on the y axis is a charged ring with the same radius as the disk and a total charge of 34.1 C. The ring is a distance d = 0.0050 m above the disk. Determine the electric field at the point P on the y axis, where P is y = 0.0100 m above the origin. FIGURE P24.35 Problems 35 and 36.arrow_forwardA conducting rod carrying a total charge of +9.00 C is bent into a semicircle of radius R = 33.0 cm, with its center of curvature at the origin (Fig.P24.75). The charge density along the rod is given by = 0 sin , where is measured clockwise from the +x axis. What is the magnitude of the electric force on a 1.00-C charged particle placed at the origin?arrow_forward
- If the curved rod in Figure P24.32 has a uniformly distributed charge Q = 35.5 nC, radius R = 0.785 m, and = 60.0, what is the magnitude of the electric field at point A?arrow_forwardA When we find the electric field due to a continuous charge distribution, we imagine slicing that source up into small pieces, finding the electric field produced by the pieces, and then integrating to find the electric field. Lets see what happens if we break a finite rod up into a small number of finite particles. Figure P24.77 shows a rod of length 2 carrying a uniform charge Q modeled as two particles of charge Q/2. The particles are at the ends of the rod. Find an expression for the electric field at point A located a distance above the midpoint of the rod using each of two methods: a. modeling the rod with just two particles and b. using the exact expression E=kQy12+y2 c. Compare your results to the exact expression for the rod by finding the ratio of the approximate expression to the exact expression. FIGURE P24.77 Problems 77 and 78.arrow_forwardThree particles with charges of 1.0 C, 1.0 C, and 0.50 C are placed at the corners A, B, and C of an equilateral triangle with side length 0.10 m as shown in Figure P23.72. Find the net force on the charge at point C.arrow_forward
- What are the magnitude and direction of a uniform electric field perpendicular to the ground that is able to suspend a particle of mass m = 2.00 g carrying a charge of +6.00 C in midair, assuming gravity and the electrostatic force are the only forces exerted on the particle?arrow_forwardA sphere with a charge of 3.50 nC and a radius of 1.00 cm is located at the origin of a coordinate system. a. What is the electric field 1.75 cm away from the center of the sphere along the positive y axis? b. If a particle with a charge of 5.39 nC were placed at that location, what would be the electrostatic force on this charge?arrow_forwardA Two positively charged particles, each with charge Q, are held at positions (a, 0) and (a, 0) as shown in Figure P23.73. A third positively charged particle with charge q is placed at (0, h). a. Find an expression for the net electric force on the third particle with charge q. b. Show that the two charges Q behave like a single charge 2Q located at the origin when the distance h is much greater than a. Figure P23.73 Problems 73 and 74.arrow_forward
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning