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Chapter 21: Problem 24

Two tiny, spherical water drops, with identical charges of \(-1.00 \times 10^{-16} \mathrm{C}\), have a center-to-center separation of \(1.00 \mathrm{~cm}\). (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

Short Answer

Expert verified
The electrostatic force is \( 8.99 \times 10^{-19} \mathrm{N} \) and there are about 625 excess electrons on each drop.

Step by step solution

01

Calculate the Electrostatic Force using Coulomb's Law

To find the electrostatic force between the two charges, we use Coulomb's Law. The formula is: \[ F = k \frac{|q_1 q_2|}{r^2} \] where:- \( F \) is the electrostatic force,- \( k = 8.99 \times 10^9 \mathrm{~N \cdot m^2/C^2} \) is Coulomb's constant,- \( q_1 = q_2 = -1.00 \times 10^{-16} \mathrm{C} \) are the charges,- \( r = 1.00 \times 10^{-2} \mathrm{m} \) is the separation distance.Substitute the values into the equation:\[ F = 8.99 \times 10^9 \times \frac{(1.00 \times 10^{-16})^2}{(1.00 \times 10^{-2})^2} \] \[ F = 8.99 \times 10^9 \times \frac{1.00 \times 10^{-32}}{1.00 \times 10^{-4}} \] \[ F = 8.99 \times 10^9 \times 1.00 \times 10^{-28} \] \[ F = 8.99 \times 10^{-19} \mathrm{N} \] The magnitude of the electrostatic force is \( 8.99 \times 10^{-19} \mathrm{N} \).
02

Calculate the Number of Excess Electrons on Each Drop

To find the number of excess electrons, use the formula:\[ n = \frac{q}{e} \] where:- \( n \) is the number of excess electrons,- \( q = -1.00 \times 10^{-16} \mathrm{C} \) is the charge on each drop,- \( e = 1.60 \times 10^{-19} \mathrm{C} \) is the elementary charge (charge of one electron).Substitute the given values:\[ n = \frac{1.00 \times 10^{-16}}{1.60 \times 10^{-19}} \] \[ n \approx 625 \] Thus, there are approximately 625 excess electrons on each drop.

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Key Concepts

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

Understanding Electrostatic Force
Electrostatic force is a fundamental concept in physics that describes the force between two charged objects. According to Coulomb's Law, this force can be calculated using the formula:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
Here, \( F \) is the electrostatic force between the objects, \( k \) is Coulomb's constant with a value of \( 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2 \), \( q_1 \) and \( q_2 \) are the magnitudes of the electric charges, and \( r \) is the distance between the centers of the two charges.
The force has the following characteristics:
  • It is attractive if the charges are opposite and repulsive if the charges are the same.
  • The force is inversely proportional to the square of the separation distance, \( r \).
  • It acts along the line joining the centers of the two charges.
Understanding these principles is crucial because they apply to various phenomena in electrostatics and beyond.
In our exercise, the magnitude of the electrostatic force, calculated using the respective charges and distance, was found to be approximately \( 8.99 \times 10^{-19} \text{ N} \). This tiny force reflects the tiny scale of the charges and the separation distance.
What is an Elementary Charge?
The elementary charge, one of the fundamental constants in physics, represents the electric charge carried by a single proton (or the negative charge of an electron). It has the value:
  • \( e = 1.60 \times 10^{-19} \text{ C} \)
This measurement is considered a basic unit of charge and plays a vital role when discussing charge quantization in atoms and molecules.
Every charge in nature is generally an integer multiple of this elementary charge, indicating that charge comes in discrete packets.
In the context of our problem, understanding the value of the elementary charge allows us to calculate the number of excess electrons, showing how charge disparity contributes to particle behavior.
Exploring Excess Electrons
Excess electrons are the additional numbers of electrons on an object, giving it a net negative charge. When discussing charged bodies, excess electrons indicate how many extra electrons an object has, compared to being neutral.
For calculating the number of excess electrons, the formula used is:
  • \( n = \frac{q}{e} \)
Where \( n \) represents the number of excess electrons, \( q \) is the total charge, and \( e \) is the elementary charge.
This formula effectively tells us how many basic charge units make up the total charge.
In our exercise, the water drops had a charge of \(-1.00 \times 10^{-16} \text{ C} \), and when calculated, each drop was found to have approximately 625 excess electrons.
Understanding excess electrons helps grasp concepts like current conduction in materials and explains why charged objects attract or repel each other.

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Most popular questions from this chapter

An electron is in a vacuum near Earth's surface and located at \(y=0\) on a vertical \(y\) axis. At what value of \(y\) should a second electron be placed such that its electrostatic force on the first electron balances the gravitational force on the first electron?

A charged nonconducting rod, with a length of \(2.00 \mathrm{~m}\) and a cross- sectional area of \(4.00 \mathrm{~cm}^{2}\), lies along the positive side of an \(x\) axis with one end at the origin. The volume charge density \(\rho\) is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if \(\rho\) is (a) uniform, with a value of \(-4.00 \mu \mathrm{C} / \mathrm{m}^{3}\), and \((\mathrm{b})\) nonuniform, with a value given by \(\rho=b x^{2}\), where \(b=-2.00 \mu \mathrm{C} / \mathrm{m}^{5} ?\)

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

A particle of charge \(+3.00 \times 10^{-6} \mathrm{C}\) is \(12.0 \mathrm{~cm}\) distant from a second particle of charge \(-1.50 \times 10^{-6} \mathrm{C} .\) Calculate the magnitude of the electrostatic force between the particles.

Two point charges of \(30 \mathrm{nC}\) and \(-40 \mathrm{nC}\) are held fixed on an \(x\) axis, at the origin and at \(x=72 \mathrm{~cm}\), respectively. A particle with a charge of \(42 \mu \mathrm{C}\) is released from rest at \(x=28 \mathrm{~cm}\). If the initial acceleration of the particle has a magnitude of \(100 \mathrm{~km} / \mathrm{s}^{2}\), what is the particle's mass?
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