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Chapter 24: Problem 17

How many electrons are contained in \(1.0 \mathrm{C}\) of charge? What is the mass of the electrons in \(1.0 \mathrm{C}\) of charge?

Short Answer

Expert verified
There are \(6.25 \times 10^{18}\) electrons in \(1.0 \mathrm{C}\) and they have a mass of \(5.69 \times 10^{-12} \mathrm{kg}\).

Step by step solution

01

Understand the electron charge

Each electron has a charge of approximately \(-1.6 \times 10^{-19} \mathrm{C}\). This is a fundamental constant of nature.
02

Calculate number of electrons

To find the number of electrons in \(1.0 \mathrm{C}\) of charge, we use the formula \(\text{number of electrons} = \frac{\text{total charge}}{\text{charge of one electron}}\). Thus, \(\text{number of electrons} = \frac{1.0 \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C/electron}} = 6.25 \times 10^{18}\).
03

Understand the electron mass

Each electron has a mass of approximately \(9.11 \times 10^{-31} \mathrm{kg}\). This is another fundamental constant of nature.
04

Calculate total mass of the electrons

Multiply the number of electrons by the mass of one electron to find the total mass. The calculation is as follows: \(\text{total mass} = 6.25 \times 10^{18} \times 9.11 \times 10^{-31} \mathrm{kg} = 5.69 \times 10^{-12} \mathrm{kg}\).

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Key Concepts

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

Number of Electrons
When dealing with electric charge, understanding the number of electrons involved is crucial. Electrons are tiny particles that carry negative charge and are a fundamental component of atoms.
Each electron carries a specific charge value, which is approximately \(-1.6 \times 10^{-19} \mathrm{C}\). This is known as the elementary charge.
To find the total number of electrons corresponding to a certain charge, such as \(1.0 \mathrm{C}\), we use the formula:
  • \( \text{number of electrons} = \frac{\text{total charge}}{\text{charge of one electron}} \)
This means for \(1.0 \mathrm{C}\), you would calculate:
  • \( \frac{1.0 \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C/electron}} = 6.25 \times 10^{18} \) electrons.
This large number illustrates just how many electrons are needed to make up just one coulomb of electric charge.
Mass of Electrons
Electrons not only carry charge but also have mass, albeit tiny. Each electron has a mass of approximately \(9.11 \times 10^{-31} \mathrm{kg}\), which is negligible on a macroscopic level but essential for precise calculations in physics.
To find the total mass of a collection of electrons, such as those found in \(1.0 \mathrm{C}\) of charge, you multiply the number of electrons by the mass of each individual electron.
For example:
  • \( \text{total mass} = 6.25 \times 10^{18} \times 9.11 \times 10^{-31} \mathrm{kg} = 5.69 \times 10^{-12} \mathrm{kg} \)
While this mass may seem small, it reflects the cumulative mass of many electrons working together.
Fundamental Constants
Certain values in physics, such as the charge and mass of an electron, are known as fundamental constants. These constants provide a quantitative foundation for scientific calculations by giving precise, objective measurements.
The elementary charge, \(1.6 \times 10^{-19} \mathrm{C}\), and the electron mass, \(9.11 \times 10^{-31} \mathrm{kg}\), are used globally to calculate quantities in both academic and applied physics.
  • These constants are derived from extensive experiments and observations, giving consistency across all measurements and theories.
  • They are critical when performing calculations involving large numbers of particles, such as determining the charge or mass of a substantial number of electrons.
  • Using these constants allows us to understand and predict the behavior of particles on an atomic level accurately.
Knowing these constants helps physicists and students alike to bridge the gap between theoretical concepts and practical applications.

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Most popular questions from this chapter

Two charged metal plates in vacuum are \(15 \mathrm{~cm}\) apart as drawn in Fig. \(24-7\). The electric field between the plates is uniform and has a strength of \(E=3000 \mathrm{~N} / \mathrm{C}\). An electron \(\left(q=-e, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right)\) is released from rest at point \(P\) just outside the negative plate. ( \(a\) ) How long will it take to reach the other plate? \((b)\) How fast will it be going just before it hits? The electric field lines show the force on a positive charge. (A positive charge would be repelled to the right by the positive plate and attracted to the right by the negative plate.) An electron, being negative, will experience a force in the opposite direction, toward the left, of magnitude $$F_{E}=|q| E=\left(1.6 \times 10^{-19} \mathrm{C}\right)(3000 \mathrm{~N} / \mathrm{C})=4.8 \times 10^{-16} \mathrm{~N}$$ Because of this force, the electron experiences an acceleration toward the left given by $$a=\frac{F_{E}}{m}=\frac{4.8 \times 10^{-16} \mathrm{~N}}{9.1 \times 10^{-31} \mathrm{~kg}}=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}$$ In the motion problem for the electron released at the negative plate and traveling to the positive plate, $$v_{i}=0 \quad x=0.15 \mathrm{~m} \quad a=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}$$ (a) From \(x=v_{i} t+\frac{1}{2} a t^{2}\) we have $$t=\sqrt{\frac{2 x}{a}}=\sqrt{\frac{(2)(0.15 \mathrm{~m})}{5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}}}=2.4 \times 10^{-8} \mathrm{~s}$$ (b) \(\quad v=v_{i}+a t=0+\left(5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\right)\left(2.4 \times 10^{-8} \mathrm{~s}\right)=1.30 \times 10^{7} \mathrm{~m} / \mathrm{s}\) As you will see in Chapter 41 , relativistic effects begin to become important at speeds above this. Therefore, this approach must be modified for very fast particles.

Three point charges in vacuum are placed on the \(x\) -axis in Fig. \(24-1 .\) Find the net force on the \(-5 \mu \mathrm{C}\) charge due to the two other charges. Because unlike charges attract, the forces on the \(-5 \mu \mathrm{C}\) charge are as shown. The magnitudes of \(\overrightarrow{\mathbf{F}}_{E 3}\) and \(\overrightarrow{\mathbf{F}}_{E 8}\) are given by Coulomb's Law: $$\begin{array}{l} F_{E 3}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left(3.0 \times 10^{-6} \mathrm{C}\right)\left(5.0 \times 10^{-6} \mathrm{C}\right)}{(0.20 \mathrm{~m})^{2}}=3.4 \mathrm{~N} \\ F_{E 8}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left(8.0 \times 10^{-6} \mathrm{C}\right)\left(5.0 \times 10^{-6} \mathrm{C}\right)}{(0.30 \mathrm{~m})^{2}}=4.0 \mathrm{~N} \end{array}$$ Keep in mind the following: (1) Proper units (coulombs and meters) must be used. (2) Because we want only the magnitudes of the forces, we do not carry along the signs of the charges. That is, we use their absolute values. Determine if the forces are attractive or repulsive and then draw them in your diagram. Pick a direction to be positive and sum the forces. From the diagram, the resultant force on the center charge is $$F_{E}=F_{E 8}-F_{E 3}=4.0 \mathrm{~N}-3.4 \mathrm{~N}=0.6 \mathrm{~N}$$ and it is in the \(+x\) -direction, to the right.

Find the ratio of the Coulomb electric force \(F_{E}\) to the gravitational force \(F_{G}\) between two electrons in vacuum. From Coulomb's Law and Newton's Law of gravitation, The electric force is much stronger than the gravitational force. $$F_{E}=k \frac{q_{\bullet}^{2}}{r^{2}} \quad \text { and } \quad F_{G}=G \frac{m^{2}}{r^{2}}$$ therefore, $$ \begin{aligned} \frac{F_{E}}{F_{G}} &=\frac{k q_{0}^{2} / r^{2}}{G m^{2} / r^{2}}=\frac{k q_{\bullet}^{2}}{G m^{2}} \\ &=\frac{\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(9.1 \times 10^{-31} \mathrm{~kg}\right)^{2}}=4.2 \times 10^{42} \end{aligned} $$

If two equal point charges, each of \(1 \mathrm{C}\), were separated in air by a distance of \(1 \mathrm{~km}\), what would be the force between them?

Two small spheres in vacuum are \(1.5 \mathrm{~m}\) apart center-to-center. They carry identical charges. Approximately how large is the charge on each if each sphere experiences a force of \(2 \mathrm{~N}\) ? The diameters of the spheres are small compared to the \(1.5 \mathrm{~m}\) separation. We may therefore approximate them as point charges. Coulomb's Law, \(F_{E}=k_{0} q_{.1} q_{.2} / r^{2}\), leads to $$q_{01} q_{\cdot 2}=q^{2}=\frac{F_{E} r^{2}}{k_{0}}=\frac{(2 \mathrm{~N})(1.5 \mathrm{~m})^{2}}{9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}=5 \times 10^{-10} \mathrm{C}^{2}$$ from which \(q=2 \times 10^{-5} \mathrm{C}\).
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