Problem 85 - University Physics with Modern Physics

Question

Two small, copper spheres each have radius 1.00 \(\mathrm{mm}\) . (a) How many atoms does each sphere contain? (b) Assume that each copper atom contains 29 protons and 29 electrons. We know that electrons and protons have charges of exactly the same magnitude, but let's explore the effect of small differences (see also Problem 21.83\()\) . If the charge of a proton is \(+e\) and the magnitude of the charge of an electron is 0.100\(\%\) smaller, what is the net charge of each sphere and what force would one sphere exert on the other if they were separated by 1.00 \(\mathrm{m} ?\)

Step-by-Step Solution

Answer

a) Each sphere contains about \( 3.55 \times 10^{20} \) atoms. b) The net charge on each sphere is approximated as 1.65 C. The force between the spheres is about \( 24.5 \times 10^9 \, \text{N} \).

Step 1: Find the Volume of the Sphere

The formula for the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \). Given the radius \( r = 1.00 \text{ mm} = 1.00 \times 10^{-3} \text{ m} \), the volume \( V \) is calculated as follows: \[ V = \frac{4}{3}\pi (1.00 \times 10^{-3})^3 \approx 4.19 \times 10^{-9} \text{ m}^3 \]

Step 2: Calculate the Number of Atoms per Sphere

Copper has a density of \( 8.96 \times 10^3 \text{ kg/m}^3 \) and an atomic mass of about \( 63.55 \text{ g/mol} \). The number of atoms in the sphere can be calculated using:1. Mass \( = \text{density} \times \text{volume} = 8.96 \times 10^3 \times 4.19 \times 10^{-9} \approx 3.75 \times 10^{-5} \text{ kg} \).2. Moles \( = \text{mass (in grams)} / \text{molar mass} = (3.75 \times 10^{-2}) / 63.55 \approx 5.90 \times 10^{-4} \text{ mol} \).3. Number of atoms \( = \text{moles} \times \text{Avogadro's number} = 5.90 \times 10^{-4} \times 6.022 \times 10^{23} \approx 3.55 \times 10^{20} \).

Step 3: Calculate the Charge Imbalance

Electrons have a charge magnitude of approximately \( -0.999e \), with a proton charge being \( +e \). The net charge per atom is \( 29e - 29(0.999e) = 0.029e \).

Step 4: Find the Net Charge of Each Sphere

With roughly \( 3.55 \times 10^{20} \) atoms in each sphere, the net charge is given by:\[ \text{Net charge} = 3.55 \times 10^{20} \times 0.029e = 1.030 \times 10^{19} e \] Given 1 electron charge \( e = 1.602 \times 10^{-19} \text{ C} \), \[ \text{Net charge in Coulombs} = 1.030 \times 10^{19} \times 1.602 \times 10^{-19} \approx 1.65 \text{ C} \].

Step 5: Calculate the Force between the Spheres

Using Coulomb's law, the force \( F \) is calculated as:\[ F = \frac{k \times |q_1| \times |q_2|}{r^2} \]where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( |q_1| = |q_2| = 1.65 \text{ C} \), and \( r = 1.00 \text{ m} \), \[ F = \frac{8.99 \times 10^9 \times (1.65)^2}{1.00^2} \approx 24.5 \times 10^9 \, \text{N} \].

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics that describes the force between two charged objects. The law states that the force (\( F \)) between two point charges is directly proportional to the product of their magnitudes (\( q_1 \) and \( q_2 \)) and inversely proportional to the square of the distance (\( r \)) between them. Mathematically, it is expressed as:

• \[ F = \frac{k \times |q_1| \times |q_2|}{r^2} \]

The constant \( k \) is known as the electrostatic constant, with a value of approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).

Coulomb's Law helps us understand how charges interact over a distance. For instance, in the problem with two copper spheres, each having a net charge, the law is used to calculate the force exerted by one sphere on the other. These calculations are crucial for predicting behavior in systems where electric forces dominate.

Charge Imbalance

A charge imbalance occurs when the number of protons (positive charges) does not equal the number of electrons (negative charges) in an object. Such an imbalance results in a net charge. In our exercise involving copper spheres, a slight difference in charge magnitude was explored, which leads to this charge imbalance.

Protons typically carry a charge of \(+e\) and electrons carry a charge of \(-0.999e\) (as given in the problem). This variation results in a small net positive charge per atom because the proton's charge slightly overpowers the electron's charge. For each copper atom, we calculate the net charge by:

• \( \text{Net charge per atom} = 29e - 29(0.999e) = 0.029e \)

Understanding charge imbalance is essential in explaining various electrical phenomena, such as the forces between charged objects and resulting electric fields.

Number of Atoms

The number of atoms in a material is a foundational concept in chemistry and physics. Knowing the number of atoms in an object allows us to determine its mass and understand its chemical properties. For a small copper sphere with a given density and atomic mass, we can compute the number of atoms using these steps:

• Calculate the mass of the sphere using density and volume: \( \text{mass} = \text{density} \times \text{volume} \).
• Convert the mass from kilograms to grams for easier use in chemistry: \( 3.75 \times 10^{-5} \text{ kg} = 3.75 \times 10^{-2} \text{ g} \).
• Find the number of moles using the atomic mass of copper: \( \text{moles} = \frac{\text{mass (in grams)}}{\text{molar mass}} \).
• Finally, calculate the number of atoms using Avogadro's number: \( \text{atoms} = \text{moles} \times 6.022 \times 10^{23} \text{ atoms/mol} \).

In this exercise, these calculations show that each sphere contains approximately \( 3.55 \times 10^{20} \) atoms. This step is critical in determining the overall charge of the sphere when coupled with charge imbalance.

Copper Sphere

Copper spheres in this context are useful for exploring basic principles of electrostatics due to copper's well-understood properties. Copper is a metal with good electrical conductivity, and it easily allows for charge accumulation and transfer. Key aspects of a copper sphere in such exercises include:

• **Volume and Density**: Understanding these allows us to calculate mass and the number of atoms.
• **Atomic Structure**: Each copper atom contains 29 protons and usually, 29 electrons — until a specified charge imbalance is considered.
• **Electrostatic Properties**: Due to its conductive nature, it's possible to explore charge distribution and interactions through models like these spheres.

In our problem, the copper sphere serves as a practical example to apply theoretical principles like Coulomb's Law, atomic theory, and charge imbalance. By examining these objects, students gain insight into how these concepts come together to describe real-world phenomena.