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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Until recently, TV sets consisted of what was called a cathode-ray tube. Electrons were boiled off a light bulb filament and passed through plates that deflected the particles; the electron beam went on to strike the TV screen itself and lit up a pixel of phosphor. The beam could be deflected both up and down and sideways (only one direction shown). By deflecting the electron beam and scanning it across the screen a picture was built up. Consider the simplified TV below. An electron beam is shot along the centerline between the two deflector plates. The top plate is held by a power supply (not shown) at a voltage \(V=\) \(+1000\) volts above the bottom plate. The separation between the plates is \(d=4 \mathrm{~cm}\). The length of the plates is \(\ell=8 \mathrm{~cm}\). The distance \(L\) from the end of the deflector plate to the TV screen is \(30 \mathrm{~cm}\). The electrons enter the region between the deflector plates with a velocity \(v=4 \times 10^7 \mathrm{~m} / \mathrm{s}\). a) In which direction is the electric field between the deflector plates? b) In which direction are the electrons deflected? c) What is the force on an electron? (Neglect gravity.) d) What is the acceleration on the electron? e) What is the total deflection of the electron as it hits the TV screen? f) What size magnetic field and in what direction would you need to place between the deflector plates in order to prevent the electrons from being deflected?

Two identical point charges \(q_1\) and \(q_2\) are at a distance \(r\) apart. If the size of \(q_1\) is doubled and the distance between them tripled, the strength of the electrical force between them (A) goes up by a factor of 3 . (B) goes down by a factor of 3 . (C) goes down by a factor of 9 . (D) goes down by a factor of \(2 / 3\). (E) goes down by a factor of \(2 / 9\)

Two identical conducting spheres, \(A\) and \(B\) carry charges \(+Q\) and \(-Q\), respectively. A third, identical conducting sphere \(C\) carries charge \(Q=0\). Sphere \(A\) is touched to sphere \(C\) and separated. Next, sphere \(B\) is touched to sphere \(C\) and separated. Finally, \(A\) is touched to \(B\) and separated. What is the final charge on each sphere? (A) \(A=Q ; B=-Q ; C=0\) (B) \(A=Q / 2 ; B=Q / 2 ; C=Q / 4\) (C) \(A=Q / 8 ; B=Q / 8 ; C=-Q / 4\) (D) \(A=Q / 2 ; B=-Q / 4 ; C=-Q / 4\) (E) \(A=Q / 4 ; B=-Q / 8 ; C=-Q / 4\)

A test charge \(+q\) and a test charge \(-q\) are released midway between the two plates. Let the voltage of the top plate be \(V\) and the voltage of the bottom plate be 0 . The distance between the two plates is \(d\). Which of the following statements about the test charges are true? i. Both charges have gained a kinetic energy of \(|q| E d / 2\) when they hit the plates. ii. Charge \(+q\) is initially at a positive voltage and charge \(-q\) is initially at a negative voltage. iii. Charge \(+q\) initially has a positive potential energy and charge \(-q\) initially has a negative potential energy. iv. Both charges lose potential energy. (A) i only (B) i and ii (C) i and iii (D) iii and iv (E) i, iii and iv

Two long, parallel wires separated by a distance \(r\) carry equal currents \(I\) in opposite directions, as shown. The direction of the field caused by the top wire at the position of the bottom wire and the direction of the force exerted by the top wire on the bottom wire are (A) \(B\) into the page; \(F\) down (B) \(B\) up; \(F\) into the page (C) \(B\) into the page; \(F\) up (D) \(B\) out of the page; \(F\) down (E) \(B\) down; \(F\) out of the page

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free