/*! 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 38 Figure \(21-37\) shows four iden... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 38

Figure \(21-37\) shows four identical conducting spheres that are actually well separated from one another. Sphere \(W\) (with an initial charge of zero) is touched to sphere \(A\) and then they are separated. Next, sphere \(W\) is touched to sphere \(B\) (with an initial charge of \(-32 e\) ) and then they are separated. Finally, sphere \(W\) is touched to sphere \(C\) (with an initial charge of \(+48 e\) ), and then they are separated. The final charge on sphere \(W\) is \(+18 e\). What was the initial charge on sphere \(A\) ?

Short Answer

Expert verified
The initial charge on sphere A was \(+16e\).

Step by step solution

01

Understanding Initial Conditions

Sphere W starts with a charge of zero, Sphere A has an unknown initial charge (let's call it \(q_A\)), Sphere B starts with \(-32e\), and Sphere C has \(+48e\). Sphere W is used to redistribute the charge among these spheres.
02

Touching Sphere W to Sphere A

When Sphere W is touched to Sphere A, they share their total charge equally. Since Sphere W initially has \(0\) and Sphere A has \(q_A\), after separation, each sphere holds \(\frac{q_A + 0}{2} = \frac{q_A}{2}\) charge.
03

Touching Sphere W to Sphere B

Now, Sphere W with \(\frac{q_A}{2}\) is touched to Sphere B with \(-32e\). They share their total charge equally, so each gets \(\frac{\frac{q_A}{2} - 32}{2}\). After separation, Sphere W has \(\frac{q_A - 64}{4}\) charge.
04

Touching Sphere W to Sphere C

Sphere W now is touched to Sphere C with \(+48e\). After touching, they share the total charge equally, making each sphere receive \(\frac{\frac{q_A - 64}{4} + 48}{2}\). After separation, Sphere W has \(\frac{q_A - 64 + 192}{8}\) which simplifies to \(\frac{q_A + 128}{8}\).
05

Solving for the Initial Charge of Sphere A

The final charge on Sphere W is known to be \(+18e\). Setting \(\frac{q_A + 128}{8} = 18e\) and solving for \(q_A\) gives: \[ q_A + 128 = 144 \] \[ q_A = 16e \]. Thus, the initial charge on Sphere A was \(+16e\).

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 91影视!

Key Concepts

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

Charge Redistribution
When we discuss charge redistribution, we're looking at how electric charge moves and balances out among different objects when they come into contact. This is a key idea in electrostatics. When two objects made of conductive material touch, charge can move from one to the other until equilibrium is reached. This means that the charge is evenly spread out across both objects.
  • If one object is neutral and the other is charged, the charge will distribute evenly, with both objects ending up with half of the total initial charge.
  • In the case of Sphere W and Sphere A in our exercise, after touching and separating, each sphere receives half of Sphere A's initial charge, since Sphere W started with zero charge.
Understanding charge redistribution is essential for solving problems where objects repeatedly come into contact and share their charge.
Conducting Spheres
Conducting spheres are objects that allow electric charge to flow freely across their surface. This ability makes them ideal for studying electrostatic phenomena, as they can easily exchange charge with other conductors.
  • Conductors, like metals, have electrons that can move about freely, unlike insulators where electrons are more tightly bound.
  • In the scenario from the exercise, the conducting spheres W, A, B, and C can exchange charge each time they are brought into contact and then separated. This process ensures that charge is distributed evenly between touching spheres.
The behavior of conducting spheres is predictable because of the uniform distribution of charge when they are symmetrically shaped and isolated from external influences.
Electric Charge Conservation
Electric charge conservation is a fundamental principle in physics. It states that the total charge in an isolated system remains constant over time, no matter the interactions that occur.
  • This principle is important because it allows us to predict and calculate the final charges on objects after they interact, like in our sphere exercise.
  • Each time Sphere W touches another sphere, the total charge before and after their interaction remains unchanged. The charge is simply redistributed among the spheres.
This concept helps us solve for unknown quantities, such as the initial charge on Sphere A, by examining how charge moves and is conserved during each contact event.

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

How many megacoulombs of positive charge are in \(1.00 \mathrm{~mol}\) of neutral molecular-hydrogen gas \(\left(\mathrm{H}_{2}\right)\) ?

Two particles are fixed on an \(x\) axis. Particle 1 of charge \(40 \mu \mathrm{C}\) is located at \(x=-2.0 \mathrm{~cm} ;\) particle 2 of charge \(Q\) is located at \(x=3.0 \mathrm{~cm} .\) Particle 3 of charge magnitude 20 \(\mu \mathrm{C}\) is released from rest on the \(y\) axis at \(y=2.0 \mathrm{~cm}\). What is the value of \(O\) if the initial acceleration of particle 3 is in the positive direction of (a) the \(x\) axis and (b) the \(y\) axis?

Of the charge \(Q\) initially on a tiny sphere, a portion \(q\) is to be transferred to a second, nearby sphere. Both spheres can be treated as particles and are fixed with a certain separation. For what value of \(q / Q\) will the electrostatic force between the two spheres be maximized?

What is the magnitude of the electrostatic force between a singly charged sodium ion \(\left(\mathrm{Na}^{+}\right.\), of charge \(+e\) ) and an adjacent singly charged chlorine ion \(\left(\mathrm{Cl}^{-}\right.\), of charge \(\left.-e\right)\) in a salt crystal if their separation is \(2.82 \times 10^{-10} \mathrm{~m}\) ?

The charges and coordinates of two charged particles held fixed in an \(x y\) plane are \(q_{1}=+3.0 \mu \mathrm{C}, x_{1}=3.5 \mathrm{~cm}, y_{1}=0.50 \mathrm{~cm}\), and \(q_{2}=-4.0 \mu \mathrm{C}, x_{2}=-2.0 \mathrm{~cm}, y_{2}=1.5 \mathrm{~cm}\). Find the (a) magni- tude and (b) direction of the electrostatic force on particle 2 due to particle 1 . At what (c) \(x\) and (d) \(y\) coordinates should a third particle of charge \(q_{3}=+4.0 \mu \mathrm{C}\) be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Fundamentals of Physics

Read Explanation

Electromagnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Solid State Physics

Read Explanation

Medical Physics

Read Explanation
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.