(II) A person scuffing her feet on a wool rug on a dry day accumulates a net charge of -28 μC. How many excess electrons does she get, and by how much does her mass increase?
(II) A person scuffing her feet on a wool rug on a dry day accumulates a net charge of -28 μC. How many excess electrons does she get, and by how much does her mass increase?
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please type out your solution so that it is easy to read I have bad eyesight
![### Problem 8
#### Electrostatics and Charge
A person scuffing her feet on a wool rug on a dry day accumulates a net charge of \(-28 \, \mu\text{C}\). To solve this problem, we need to determine two things:
1. **The number of excess electrons she gets:**
Using the elementary charge of an electron, \(e = 1.602 \times 10^{-19} \, \text{C}\), we can find the number of excess electrons.
\[
\text{Number of excess electrons} = \frac{\text{Net Charge}}{\text{Charge of one electron}} = \frac{28 \times 10^{-6} \, \text{C}}{1.602 \times 10^{-19} \, \text{C/electron}}
\]
2. **The increase in her mass due to the accumulated charge:**
The mass of one electron is approximately \(9.109 \times 10^{-31} \, \text{kg}\). Once we have the number of excess electrons, we can compute the total mass increase.
##### Detailed Calculation
1. Calculate the number of excess electrons:
\[
N = \frac{28 \times 10^{-6} \, \text{C}}{1.602 \times 10^{-19} \, \text{C/electron}} \approx 1.75 \times 10^{14} \, \text{electrons}
\]
2. Calculate the mass increase:
\[
\Delta m = N \times \text{mass of one electron} = 1.75 \times 10^{14} \times 9.109 \times 10^{-31} \, \text{kg} \approx 1.59 \times 10^{-16} \, \text{kg}
\]
Thus, she gets approximately \(1.75 \times 10^{14}\) excess electrons, and her mass increases by about \(1.59 \times 10^{-16} \, \text{kg}\).
This example illustrates the practical application of fundamental physical constants in understanding how charge and mass interact at a microscopic level.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcaea1532-9bca-456b-84fc-9594bcc8dc11%2Fef49957b-2e58-4d79-b5c4-538aedf0471d%2Fzg1m1gq_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 8
#### Electrostatics and Charge
A person scuffing her feet on a wool rug on a dry day accumulates a net charge of \(-28 \, \mu\text{C}\). To solve this problem, we need to determine two things:
1. **The number of excess electrons she gets:**
Using the elementary charge of an electron, \(e = 1.602 \times 10^{-19} \, \text{C}\), we can find the number of excess electrons.
\[
\text{Number of excess electrons} = \frac{\text{Net Charge}}{\text{Charge of one electron}} = \frac{28 \times 10^{-6} \, \text{C}}{1.602 \times 10^{-19} \, \text{C/electron}}
\]
2. **The increase in her mass due to the accumulated charge:**
The mass of one electron is approximately \(9.109 \times 10^{-31} \, \text{kg}\). Once we have the number of excess electrons, we can compute the total mass increase.
##### Detailed Calculation
1. Calculate the number of excess electrons:
\[
N = \frac{28 \times 10^{-6} \, \text{C}}{1.602 \times 10^{-19} \, \text{C/electron}} \approx 1.75 \times 10^{14} \, \text{electrons}
\]
2. Calculate the mass increase:
\[
\Delta m = N \times \text{mass of one electron} = 1.75 \times 10^{14} \times 9.109 \times 10^{-31} \, \text{kg} \approx 1.59 \times 10^{-16} \, \text{kg}
\]
Thus, she gets approximately \(1.75 \times 10^{14}\) excess electrons, and her mass increases by about \(1.59 \times 10^{-16} \, \text{kg}\).
This example illustrates the practical application of fundamental physical constants in understanding how charge and mass interact at a microscopic level.
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