/*! 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 4 Four identical metal spheres hav... [FREE SOLUTION] | 91Ó°ÊÓ

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Four identical metal spheres have charges of \(q_{\mathrm{A}}=-8.0 \mu \mathrm{C}\), \(q_{\mathrm{B}}=-2.0 \mu \mathrm{C}, q_{\mathrm{C}}=+5.0 \mu \mathrm{C},\) 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 \(+3.0 \mu \mathrm{C} ?(\mathrm{c})\) 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 B, C, and D. (c) Approximately 1.875 x 10^13 electrons needed.

Step by step solution

01

Calculating combined charge for part (a)

To find two spheres that, when brought together and separated, result in each having a charge of \(+5.0 \mu \mathrm{C}\), we start by considering the initial charges. Suppose two spheres, like A and C, are brought together: \(q_\mathrm{A} = -8.0 \mu \mathrm{C}\) and \(q_\mathrm{C} = +5.0 \mu \mathrm{C}\). Their total charge is \(-8.0 \mu \mathrm{C} + 5.0 \mu \mathrm{C} = -3.0 \mu \mathrm{C}\). Splitting evenly leaves \(-1.5 \mu \mathrm{C}\) each, so this pairing does not work. Correct this process with different combinations, verify A and D: \(q_\mathrm{A} = -8.0 \mu \mathrm{C}\) and \(q_\mathrm{D} = +12.0 \mu \mathrm{C}\). Total charge: \(+4.0 \mu \mathrm{C}\), split to \(+2.0 \mu \mathrm{C}\) each, doesn't work. Try B and D: \(q_\mathrm{B} = -2.0 \mu \mathrm{C}\) and \(q_\mathrm{D} = +12.0 \mu \mathrm{C}\). Total charge: \(+10.0 \mu \mathrm{C}\), leading to \(+5.0 \mu \mathrm{C}\) each, working for part (a).
02

Arranging 3 spheres for part (b)

We want three spheres to collectively have a total charge such that splitting it evenly results in each having \(+3.0 \mu \mathrm{C}\). The combination of charges \(\{q_\mathrm{A}, q_\mathrm{B}, q_\mathrm{C}\}\) totals \(-8.0 \mu \mathrm{C} -2.0 \mu \mathrm{C} +5.0 \mu \mathrm{C} = -5.0 \mu \mathrm{C}\), not working out evenly to \(+3.0 \mu \mathrm{C}\). Try \(\{q_\mathrm{B}, q_\mathrm{C}, q_\mathrm{D}\}\): total = \(-2.0 \mu \mathrm{C} + 5.0 \mu \mathrm{C} + 12.0 \mu \mathrm{C} = 15.0 \mu \mathrm{C}\); dividing gives each \(3.0 \mu \mathrm{C}\), confirming the correct selection.
03

Determining electrons needed for neutralization in part (c)

Each sphere ends with \(+3.0 \mu \mathrm{C}\). To be neutral, they need no net charge. Since they have \(+3.0 \mu \mathrm{C}\), they need an equivalent negative charge of \(-3.0 \mu \mathrm{C}\). Calculate number of electrons required: \(3.0 \mu \mathrm{C} = 3.0 \times 10^{-6} \mathrm{C}\). Each electron has a charge of \(-1.6 \times 10^{-19} \mathrm{C}\). Therefore, the number of electrons required: \(\frac{3.0 \times 10^{-6} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C/e^-}} = 1.875 \times 10^{13}\) electrons.

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Key Concepts

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

Charge Distribution
Charge distribution is a fundamental concept in electrostatics. When we talk about charge distribution, we are referring to how electric charge is spread out over an object or between multiple objects. In electrostatics, the way charges distribute themselves when multiple charged objects come into contact can drastically influence their subsequent interactions.

For instance, in our exercise, the interaction between two metal spheres involves the redistribution of charges when they are brought into contact. Here’s how it works: when two conductors with different initial charges touch, the total combined charge distributes evenly between them after they are separated. This is why correctly pairing two spheres such as sphere B with \(q_\mathrm{B} = -2.0 \, \mu \mathrm{C}\) and sphere D with \(q_\mathrm{D} = +12.0 \, \mu \mathrm{C}\) yields the correct final charge of \(+5.0 \, \mu \mathrm{C}\) on each. Their total charge is\(+10.0 \, \mu \mathrm{C}\), which is then evenly split into \(+5.0 \, \mu \mathrm{C}\) per sphere.

This principle of even distribution is crucial to understanding charge equalization in electrostatic interactions.
Electric Charge
Electric charge is a property of subatomic particles that leads to electromagnetic interaction. Simply put, it's what causes objects to attract or repel each other. Charges come in two types: positive and negative, analogous to two groups, where like charges repel and unlike charges attract each other.

Understanding electric charges is essential in solving problems like the one in our exercise. The charges, measured in coulombs (\(\mathrm{C}\)), result from an imbalance between the number of protons and electrons in an atom. Objects can get charged by gaining or losing electrons.

In electrostatics exercises, you'll encounter calculations involving the sum of different charges, such as finding the spheres \(q_\mathrm{B} = -2.0 \, \mu \mathrm{C}\), \(q_\mathrm{C} = +5.0 \, \mu \mathrm{C}\), and\( q_\mathrm{D} = +12.0 \, \mu \mathrm{C}\). Their net charge determines how they interact when they come into contact and how charge distribution will ultimately balance out.

These calculations also define the principles governing how excess or deficit electrons on a surface determine its electric charge. This background helps us appreciate how the attractive and repulsive forces at play form the foundation of electricity and magnetism.
Neutralization
Neutralization is the process where a charged body becomes electrically neutral. This occurs when a body gains or loses electrons such that its net charge equals zero. Neutralization helps explain how electrostatic interactions often lead to electrical balance, even when initial charges start off imbalanced.

For instance, in part (c) of our example, each sphere ends with \(+3.0 \, \mu \mathrm{C}\). To neutralize this charge, each sphere must acquire an equivalent negative charge. This is achieved by adding electrons. Since the charge of a single electron is \(-1.6 \times 10^{-19} \, \mathrm{C}\), calculating the number of electrons needed requires simple division:\[ \frac{3.0 \times 10^{-6} \, \mathrm{C}}{1.6 \times 10^{-19} \, \mathrm{C/e^-}} = 1.875 \times 10^{13} \] electrons are needed to make the sphere neutral.

This calculation shows how electrons help balance out the positive charge, bringing the sphere closer to electrical neutrality. Neutralization is fundamental in various applications, from grounding electrical circuits to the operation of lightning rods.

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Most popular questions from this chapter

The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of \(7.1 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) (a) What is the magnitude of the electric field within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion \(\left(\mathrm{K}^{+} ;\right.\) charge \(\left.=+e\right)\) placed inside the membrane.

The masses of the earth and moon are \(5.98 \times 10^{24}\) and \(7.35 \times 10^{22} \mathrm{kg}\) respectively. Identical amounts of charge are placed on each body, such that the net force (gravitational plus electrical) on each is zero. What is the magnitude of the charge placed on each body?

mmh A solid nonconducting sphere has a positive charge \(q\) spread uniformly throughout its volume. The charge density or charge per unit volume, therefore, is \(\frac{q}{\frac{4}{3} \pi R^{3}} .\) Use Gauss' law to show that the electric field at a point within the sphere at a radius \(r\) has a magnitude of \(\frac{q r}{4 \pi \epsilon_{0} R^{3}}\) (Hint: For a Gaussian surface, use a sphere of radius r centered within the solid sphere of radius. Note that the net charge within any volume is the charge density times the volume.)

A charge of \(-3.00 \mu C\) is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass, which has a radius of \(0.100 \mathrm{m} .\) The charges on the circle are \(-4.00 \mu \mathrm{C}\) at the position due north and \(+5.00 \mu \mathrm{C}\) at the position due east. What are the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east.

In a vacuum, a proton (charge \(=+e,\) mass \(=1.67 \times 10^{-27} \mathrm{kg}\) ) is moving parallel to a uniform electric field that is directed along the \(+x\) axis (see the figure). The proton starts with a velocity of \(+2.5 \times 10^{4} \mathrm{m} / \mathrm{s}\) and accelerates in the same direction as the electric field, which has a value of \(+2.3 \times 10^{3} \mathrm{N} / \mathrm{C} .\) Find the velocity of the proton when its displacement is \(+2.0 \mathrm{mm}\) from the starting point.

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