/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Two masses of \(10^{-9} \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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Two masses of \(10^{-9} \mathrm{~kg}\) each carry a charge of \(+3 \mathrm{nC}\) and are initially held motionless by a massless thread at a distance of \(9 \mathrm{~m}\). a) Draw the electric field produced by the two charges. b) What is the tension in the thread? c) The thread is cut by a magic wand. What is the initial acceleration of each of the charges? d) How does the acceleration change with \(r\) ? Graph the behavior. Explain qualitatively how the velocity will change with \(r\). In particular, what happens to the velocity when \(r\) goes to infinity. Where is the maximum velocity?

Short Answer

Expert verified
Answer: The initial acceleration of each charge is \(10^6\mathrm{~\dfrac{m}{s^2}}\).

Step by step solution

01

Electric field produced by the two charges

Allow the student to draw an expected electric field diagram based on prior knowledge. The electric fields due to both charges should be directed away from the charges, as they are positively charged. b)
02

Tension in the thread

First, we need to find the electric force between the two charges using Coulomb's law: \(F_e = \dfrac{kq_1q_2}{r^2}\) Where \(F_e\) is the electric force, \(k = 8.99 \times 10^9\mathrm{~N m^2 C^{-2}}\) is the electrostatic constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them. Plugging in the given values: \(F_e = \dfrac{8.99 \times 10^9 (3 \times 10^{-9})(3 \times 10^{-9})}{(9)^2} = 0.001\mathrm{~N}\) The tension in the thread must balance the electric force, so: \(T = F_e\) Thus, the tension in the thread is \(0.001\mathrm{~N}\). c)
03

Initial acceleration of each charge

To find the initial acceleration of each charge, we can use Newton's second law of motion: \(F = ma\) Rearranging for acceleration and substituting for the electric force, we get: \(a = \dfrac{F_e}{m} = \dfrac{0.001\mathrm{~N}}{10^{-9}\mathrm{kg}} = 10^6\mathrm{~\dfrac{m}{s^2}}\) The initial acceleration of each charge is \(10^6\mathrm{~\dfrac{m}{s^2}}\). d)
04

Acceleration and velocity as a function of distance

As the charges move away from each other, as the distance \(r\) increases, the electric force between them will decrease according to Coulomb's law. As the force decreases, the acceleration decreases as well, i.e., \(a \propto \frac{1}{r^2}\). We can thus see that the acceleration is inversely proportional to the square of the distance between the charges. As for the velocity, when the charges are initially released from rest, they will start to gain some speed due to the electric force, and the velocity will increase as the forces acting on them decrease with increasing distance. As the distance goes to infinity, the electric force goes to zero, leading to zero acceleration, and the charges will reach the maximum velocity at this point. Then, a student should draw a graph showing the relationship between acceleration and distance, where the acceleration decreases with \(r^2\) and the velocity increases initially and then levels off.

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Key Concepts

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

Electric Field
Imagine you're standing in a field holding two small magnets. If you release one, it will start moving because of the magnetic force. Electric fields work in a similar way, but with electrical charges. In physics, the electric field around a charge is the force it exerts on other charges in its surroundings. It's invisible, but we can imagine it as lines radiating from a charge. For two positive charges, the field lines go outward, repelling each other, just as the charges in our exercise do.

The strength of this field is determined by the amount of charge and the distance from it. The closer you are, the stronger the field, which diminishes as you move away. By understanding electric fields, students can predict how charges will move and interact, leading to more complex phenomena like circuits and electromagnetism.
Tension in Physics
Tension is like a game of tug-of-war at the molecular level. It's the pulling force transmitted through a string, cable, or any object that can be stretched. In our problem with the two charges held by a thread, tension is what's keeping them in place against the electric repulsion. It's the hero holding two enemies apart!

In physics, it's crucial to recognize tension not just as a force, but as one that always pulls along the line of the object. For the massless thread in the exercise, the tension equals the electric force because it's the only force keeping the thread taut and the charges stationary. When students grasp this concept, they can better understand structures like bridges, cable cars, and even the forces in muscles.
Newton's Second Law of Motion
Newton's second law is about the interaction between force, mass, and acceleration. It tells us that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. It's the why behind a bowling ball and a beach ball behaving differently when you push them with the same strength.

In the given problem, when the thread is cut, the only force acting on each charge is the electrical force. Using Newton's second law, we can determine the initial acceleration. For our tiny charged masses, this results in a surprisingly large acceleration. This law is a critical part of understanding not just our homework problem, but essentially everything that moves, from cars to planets.
Acceleration and Velocity Relationship
Velocity tells us how fast an object is moving and in what direction, while acceleration is about how the velocity changes over time. They're like siblings with a constant competition: velocity can't change without acceleration. In our magical physics world, when the thread is cut, the charges start from rest with a high acceleration. However, as they move away from each other, and the distance increases, the acceleration decreases.

The relationship is inversely proportional to the square of the distance. So, the farther apart the charges are, the less they accelerate until eventually, they're just coasting along without speeding up or slowing down. This maximum speed is reached when the electric forces can no longer accelerate them, practically when they are infinitely far apart. Visualizing this helps students understand motion not only in charged particles but also in everyday objects like cars easing to a stop or rockets blasting into space.

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Most popular questions from this chapter

A circular loop rotates around a horizontal axis coming out of the page (side view shown, below). The loop is rotating in a uniform \(B\)-field, pointed downward. For every two full rotations, how often does the induced current change direction? (A) Twice (B) Four times (C) Eight times (D) Twelve time (E) Sixteen times

What potential difference between the two plates would be needed to accelerate a hydrogen ion (a proton) from rest to a speed of \(10^6 \mathrm{~m} / \mathrm{s}\) ? (A) \(100 \mathrm{~V}\) (B) \(500 \mathrm{~V}\) (C) \(1000 \mathrm{~V}\) (D) \(5000 \mathrm{~V}\) (E) \(10,000 \mathrm{~V}\)

A solid copper sphere of radius \(R=10 \mathrm{~cm}\) carries a charge of \(+5 \mathrm{nC}\). a) Give an algebraic expression for the electric field at all distances \(r

Instead of a meter, a device (such as a battery) is attached to the coil that can generate a current. When the device is turned on the bar magnet will (A) remain at rest. (B) be attracted to the coil. (C) be repelled from the coil. (D) shoot through the coil. (E) (B) or (C) depending on which way the current is flowing in the coil

Two identical plastic balls of mass \(10 \mathrm{gm}\) each are hung by threads with length \(30 \mathrm{~cm}\) from a common point, as shown below. The balls are each charged with the same charge \(q\) and repel each other until they come to rest with a horizontal separation of \(30 \mathrm{~cm}\). a) Sketch the electric field produced by the two balls. b) Draw the force vectors on the right-hand ball. c) What is the charge \(q\) in each ball?

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