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Chapter 18: Problem 4

Four identical metal spheres have charges of \(q_{\Lambda}=-8.0 \mu \mathrm{C}, q_{\mathrm{B}}=\) \(-2.0 \mu \mathrm{C}, q_{\mathrm{c}}=+5.0 \mu \mathrm{C},\) and \(q_{\mathrm{D}}=+12.0 \mu \mathrm{C}\) (a) Two of the spheres are brought together so they touch, and then they are separated. Which spheres are they, if the final charge on each one is \(+5.0 \mu \mathrm{C} ?\) (b) In a similar manner, which three spheres are brought together and then separated, if the final charge on each of the three is \(+3.0 \mu \mathrm{C} ?\) (c) The final charge on each of the three separated spheres in part (b) is \(+3.0 \mu \mathrm{C} .\) How many electrons would have to be added to one of these spheres to make it electrically neutral?

Short Answer

Expert verified
(a) Spheres B and D, (b) Spheres A, C, and D, (c) Add \(1.875 \times 10^{13}\) electrons.

Step by step solution

01

Understanding Charge Redistribution

When two identical spheres with charges are brought together, the total charge is equally distributed between them. This is because they are identical and have the same capacitance.
02

Calculating Combined Charge for Two Spheres

To find two spheres that can result in a final charge of \(+5.0 \mu \mathrm{C}\) each after touching, we need their total initial charge to be \(2 \times 5.0 \mu \mathrm{C} = 10.0 \mu \mathrm{C}\). This total charge needs to be divided between them. Checking given charges: \(q_B = -2.0 \mu \mathrm{C}\) and \(q_D = +12.0 \mu \mathrm{C}\) give us \(-2.0 + 12.0 = 10.0 \mu \mathrm{C}\). Thus, when \(q_B\) and \(q_D\) are brought together and separated, each will have \(+5.0 \mu \mathrm{C}\).
03

Verifying Three Spheres for Equal Charge Distribution

We need to find three spheres that can have a final charge of \(+3.0 \mu \mathrm{C}\) each. This means the total initial charge should be \(3 \times 3.0 \mu \mathrm{C} = 9.0 \mu \mathrm{C}\). Checking given charges: \(q_A = -8.0 \mu \mathrm{C}\), \(q_B = -2.0 \mu \mathrm{C}\), and \(q_C = +5.0 \mu \mathrm{C}\) give us \(-8.0 + (-2.0) + 5.0 = -5.0 \mu \mathrm{C}\), not matching \(+9.0 \mu \mathrm{C}\). Instead, try \(q_A = -8.0 \mu \mathrm{C}\), \(q_C = +5.0 \mu \mathrm{C}\), and \(q_D = +12.0 \mu \mathrm{C}\). This new combination gives \(-8.0 + 5.0 + 12.0 = 9.0 \mu \mathrm{C}\), which is the correct total.
04

Calculating Electrons for Neutralization

To neutralize a sphere with \(+3.0 \mu \mathrm{C}\), we need to add electrons equal in charge. The charge of an electron is \(-1.6 \times 10^{-19}\) C, so: \( \dfrac{3.0 \times 10^{-6}\;\mathrm{C}}{1.6 \times 10^{-19}\;\mathrm{C/electron}} = 1.875 \times 10^{13}\) electrons need to be added.

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Key Concepts

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

Charge Redistribution
When dealing with electrostatics in metallic spheres, charge redistribution is a fundamental concept. Imagine two charged spheres coming into contact with each other. At the moment they touch, their individual charges combine, and then, due to their identical nature, average out once they are separated again.

Key points about charge redistribution include:
  • The total charge is conserved; it doesn't vanish or appear out of nowhere.
  • After contact, the combined charge of both spheres gets equally divided as they are identical in material and size.
  • This division is uniform due to identical capacitance between them, meaning their ability to store charge is the same.
Knowing this helps predict what charge each sphere will have after separation, improving understanding of electrostatic interactions.
Identical Spheres
In physics problems like these, the focus on identical spheres is crucial. Identical spheres mean they have the same size, shape, and are made of the same material. This uniformity ensures that when they share or redistribute charge, they do so equally.

The term "identical" simplifies calculations because:
  • They have equal capacity to hold charge ( C) due to being identical, making the problem simpler.
  • When brought together and separated, they share the charge equally because they have the same surface area and volume.
  • The symmetry in their properties takes away the complexity of differentiating individual capacitance or material properties, focusing solely on initial and final charges.
This clarity allows us to predict outcomes based solely on initial conditions and the process of charge sharing or neutralization.
Electron Charge
Understanding the concept of electron charge is critical when it comes to analyzing electrostatic problems. An electron, which is a fundamental subatomic particle, carries a negative charge of approximately \(-1.6 \times 10^{-19}\) coulombs.

In practical situations:
  • This tiny charge is the unit of measure used to discuss and calculate the behavior of charges in materials.
  • Because of its negativity, adding electrons to an object reduces its overall positive charge or increases negative charge.
  • The magnitude of the electron charge is precise, enabling accurate computation in physics problems involving charge neutralization or redistribution.
Knowing the electron charge helps us calculate how many electrons need to be added or removed to achieve a desired level of neutrality or specific charge level in study cases.
Neutralization of Charge
Neutralization of charge involves making an object electrically neutral, which means it has no net charge. In the context of the given problem, neutralizing a positively charged sphere entails adding enough electrons to counterbalance the positive charge precisely.

A few key ideas include:
  • The process involves calculating the exact number of electrons needed based on their charge, using the formula for charge, \(q = ne\), where \(n\) is the number of electrons and \(e\) is the electron's charge.
  • For a sphere with +3.0 渭C, the number of needed electrons is determined by: \(\frac{3.0 \times 10^{-6}\, \text{C}}{1.6 \times 10^{-19}\, \text{C/electron}}\).
  • By performing this computation, we derive how many electrons need to be added to bring the charge to zero exactly, depicting the effectiveness of charge neutralization.
Mastering this concept ensures a firm grasp of how charge balance occurs through electron transfer, a central theme in electrostatics.

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Most popular questions from this chapter

Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. (a) The object has a charge of \(+20.0 \mu \mathrm{C}\) and the instrument indicates that the electric force exerted on it is \(40.0 \mu \mathrm{N}\), due east. What are the magnitude and direction of the electric field? (b) What are the magnitude and direction of the electric field if the object has a charge of \(-10.0 \mu \mathrm{C}\) and the instrument indicates that the force is \(20.0 \mu \mathrm{N},\) due west?

Two identical small insulating balls are suspended by separate \(0.25-\mathrm{m}\) threads that are attached to a common point on the ceiling. Each ball has a mass of \(8.0 \times 10^{-4} \mathrm{kg} .\) Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of \(36^{\circ}\) between the threads. Determine (a) the charge on each ball and (b) the tension in the threads.

A uniform electric field exists everywhere in the \(x, y\) plane. This electric field has a magnitude of \(4500 \mathrm{N} / \mathrm{C}\) and is directed in the positive \(x\) direction. A point charge \(-8.0 \times 10^{-9} \mathrm{C}\) is placed at the origin. Determine the magnitude of the net electric field at (a) \(x=-0.15 \mathrm{m}\) (b) \(x=+0.15 \mathrm{m}\) and \((\mathrm{c}) \mathrm{y}=+0.15 \mathrm{m}\)

You and your team are designing a device that can be used to position a small, plastic object in the region between the plates of a parallel-plate capacitor. A small plastic sphere of mass \(m=1.20 \times 10^{-2} \mathrm{kg}\) carries a charge \(q=+0.200 \mu \mathrm{C}\) and hangs vertically (along the \(y\) direction) from a massless, insulating thread (length \(l=10.0 \mathrm{cm})\) between two vertical capacitor plates. When there is no electric field, the object resides at the midpoint between the plates (at \(x=0\) ). However, when there is a field between plates (in the \(\pm x\) direction) the object moves to a new equilibrium position. (a) To what value should you set the field if you want the object to be located at \(x=2.10 \mathrm{cm} ?\) (b) To what value should you set the field if you want the object to be located at \(x=-3.30 \mathrm{cm} ?\)

A small object has a mass of \(3.0 \times 10^{-3} \mathrm{kg}\) and a charge of \(-34 \mu \mathrm{C} .\) It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of \(2.5 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\) in the direction of the \(+x\) axis. Determine the magnitude and direction of the electric field.
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