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Two small aluminum spheres, each having mass \(0.0250 \mathrm{~kg}\), are separated by \(80.0 \mathrm{~cm}\). (a) How many electrons does each sphere contain? (The atomic mass of aluminum is \(26.982 \mathrm{~g} / \mathrm{mol}\), and its atomic number is \(13 .\) ) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude \(1.00 \times 10^{4} \mathrm{~N}\) (roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

Short Answer

Expert verified
The total number of electrons is determined in Steps 1 and 2. Later, the number of electrons that need to be transferred to create a certain force is calculated in Step 3. Finally, the fraction of total elections this represents is determined in Step 4.

Step by step solution

01

Calculate the Number of Atoms in One Sphere

The first part of the exercise is to find how many atoms are in one sphere. This can be accomplished by converting the mass of a sphere from kilograms to grams (since the atomic mass is given in grams), then using the atomic mass of aluminum to calculate the number of moles, and finally using Avogadro's law (\(6.02214076 × 10^{23} atoms/mol\)) to find the number of atoms. Let's denote the number of atoms by \(N\). Therefore, \[N = \frac{(0.025 kg × 10^{3} g/kg)}{26.982 g/mol} × 6.022 × 10^{23} atoms/mol\]
02

Calculate the Total Number of Electrons

The atomic number of aluminum is 13, which indicates that each aluminum atom has 13 electrons. Therefore, the total number of electrons \(E\) in one sphere is the product of the number of atoms and the atomic number, or \(E = N × 13\). Flat this equation to get our answer.
03

Determine the Number of Electrons to Transfer

Next, we need to find the number of electrons (\(n\)) that need to be transferred from one sphere to another for the spheres to exert an attractive force of \(1.00 × 10^{4} N\). This can be computed with the use of Coulomb's law: \[F = k \frac{|q1 × q2|}{r^2}\] Here, \(F\) denotes the force, \(k\) the Coulomb's constant (\(8.99 × 10^{9} N m²/C²\)), \(q1\) and \(q2\) the charges, and \(r\) the distance between the charges. Since the spheres are treated as point charges and \(n\) electrons are transferred from one sphere to the other, one sphere has a charge of \(+ne\) (where \(e\) is the charge of an electron, \(1.60 × 10^{-19} C\)) and the other has a charge of \(-ne\). By substituting these charges into Coulomb's law, we get \[F = k \frac{|ne × (-ne)|}{r^2}.\] Thus, after rearranging and solving for \(n\), we get \[n = \sqrt{\frac{Fr^2}{ke^2}}.\] By substitifying \(F\), \(r\), \(k\), and \(e\), we can find \(n\).
04

Calculate the Fraction of Transferred Electrons

To find out what fraction of all electrons this represents, divide \(n\) by the total number of electrons \(E\). This gives us the desired fraction.

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Key Concepts

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

Atomic Structure
Understanding the atomic structure is crucial for calculating the number of electrons in an object. The atomic structure refers to the arrangement of protons, neutrons, and electrons within an atom. In each atom, the core, or nucleus, contains protons and neutrons. Surrounding this nucleus are electrons that occupy certain regions called electron shells or clouds.
Atoms are distinguished by their atomic number, which indicates the number of protons in the nucleus. For example, aluminum has an atomic number of 13, meaning each aluminum atom contains 13 protons and, therefore, 13 electrons in a neutral state. Remember, the number of electrons equals the number of protons in a neutral atom.
To calculate the number of atoms in an object, you can use the object's mass, and convert it using the atomic mass and Avogadro's number. This calculation gives you the number of atomic units, which you can then use to determine the number of electrons each atom contributes to the total.
Electron Transfer
Electron transfer refers to the movement of electrons from one entity to another, creating ions or charged objects. This process is fundamental in many chemical reactions and in generating electricity. In physics problems, especially those involving forces between charged bodies, the understanding of electron transfer is vital.
When electrons are transferred, one sphere may become positively charged (losing electrons) and the other negatively charged (gaining electrons). The total number of electrons transferred can be determined by Coulomb's law if the resultant force is known. This transfer changes the charge of the objects involved, which can be calculated knowing the charge of a single electron (approximately \(1.60 × 10^{-19} C\)).
  • Initiate with knowing the total force needed.
  • Use Coulomb's law to calculate the number of electrons involved in this transfer.
  • The charge on each sphere will depend on how many electrons it gained or lost.
Coulomb's Force
Coulomb's force is the electrostatic force between two charged objects. It is one of the fundamental forces that govern the interactions between charged entities. The formula for Coulomb's force is \(F = k \frac{|q1 \times q2|}{r^2}\), showing how the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.
This equation allows you to calculate the force between two point charges when their charges and the separation between them is known. In the context of our aluminum spheres, we can treat each sphere as a point charge because of their small size compared to the large distance.
  • Basic components that govern the force include:
  • \(k\): Coulomb's constant, \(8.99 \times 10^{9} \mathrm{~Nm}^2/ ext{C}^2\).
  • \(q1\), \(q2\): the values of the charges.
  • \(r\): Distance between the centers of the charges.
This force calculation ties into understanding how many electrons need to be transferred for the spheres to exhibit a specific force.
Point Charges
Point charges simplify the model of charged objects by assuming that all the charge of an object is concentrated at a single point. This simplification is especially useful in electrostatics for calculating electric forces and fields.
When dealing with small objects like the aluminum spheres mentioned, the point charge assumption is valid if the distance between them is large compared to their size. This simplifies calculations without significant loss of accuracy.
In our problem, treating the aluminum spheres as point charges allows the use of Coulomb's law more straightforwardly, helping us understand the attraction caused by the electrons transferred. This model reduces computation because you deal only with magnitudes rather than distribution over a surface. The charge distribution is assumed even, making it easier to predict electrostatic interactions.

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Most popular questions from this chapter

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

A thin disk with a circular hole at its center, called an \(a n-\) nulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. \(\mathbf{P 2 1 . 8 7}\) ). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z\) plane, with its center at the origin. For an arbitrary point on the \(x\) -axis (the axis of the annulus), find the magnitude and direction of the electric field \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus. (c) Show that at points on the \(x\) -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\) -axis (but cannot move off the axis). The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

Point charge \(A\) is on the \(x\) -axis at \(x=-3.00 \mathrm{~cm}\). At \(x=1.00 \mathrm{~cm}\) on the \(x\) -axis its electric field is \(2700 \mathrm{~N} / \mathrm{C}\). Point charge \(B\) is also on the \(x\) -axis, at \(x=5.00 \mathrm{~cm}\). The absolute magnitude of charge \(B\) is twice that of \(A .\) Find the magnitude and direction of the total electric field at the origin if (a) both \(A\) and \(B\) are positive; (b) both are negative; (c) \(A\) is positive and \(B\) is negative; (d) \(A\) is negative and \(B\) is positive.

Imagine two \(1.0 \mathrm{~g}\) bags of protons, one at the earth's north pole and the other at the south pole. (a) How many protons are in each bag? (b) Calculate the gravitational attraction and the electric repulsion that each bag exerts on the other. (c) Are the forces in part (b) large enough for you to feel if you were holding one of the bags?

Two point charges are placed on the \(x\) -axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{~m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{~m} .\) What are the magnitude and direction of the total force exerted by these two charges on a negative point charge \(q_{3}=-6.00 \mathrm{nC}\) that is placed at the origin?

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