(I) What is the magnitude of the electric force of attraction between an iron nucleus (q= +26e) and its innermost electron if the distance between them is 1.5 × 10-¹2 m?
(I) What is the magnitude of the electric force of attraction between an iron nucleus (q= +26e) and its innermost electron if the distance between them is 1.5 × 10-¹2 m?
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![## Physics Problem Example: Electric Force of Attraction
**Problem Statement:**
1. **What is the magnitude of the electric force of attraction between an iron nucleus (\( q = +26e \)) and its innermost electron if the distance between them is \( 1.5 \times 10^{-12} \) meters?**
**Solution:**
To find the magnitude of the electric force of attraction between the iron nucleus and its innermost electron, we can use Coulomb's Law, which is given by:
\[ F = \frac{k_e \cdot |q_1 \cdot q_2|}{r^2} \]
where:
- \( F \) is the electric force,
- \( k_e \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the charges of the particles,
- \( r \) is the distance between the charges.
Given:
- The charge of the iron nucleus, \( q = +26e \),
- The charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \),
- The distance, \( r = 1.5 \times 10^{-12} \, \text{m} \).
Substituting the values:
- \( q_1 = +26e = 26 \times 1.6 \times 10^{-19} \, \text{C} \)
- \( q_2 = -e = -1.6 \times 10^{-19} \, \text{C} \)
Therefore:
\[ F = \frac{8.99 \times 10^9 \cdot |(26 \cdot 1.6 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})|}{(1.5 \times 10^{-12})^2} \]
Calculating inside the absolute value:
\[ |(26 \cdot 1.6 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})| = |(41.6 \times 10^{-19}) \cdot (-1.6 \times 10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcaea1532-9bca-456b-84fc-9594bcc8dc11%2Faf89f68a-f1a0-405f-8762-fa06756922f9%2F74uyuh_processed.png&w=3840&q=75)
Transcribed Image Text:## Physics Problem Example: Electric Force of Attraction
**Problem Statement:**
1. **What is the magnitude of the electric force of attraction between an iron nucleus (\( q = +26e \)) and its innermost electron if the distance between them is \( 1.5 \times 10^{-12} \) meters?**
**Solution:**
To find the magnitude of the electric force of attraction between the iron nucleus and its innermost electron, we can use Coulomb's Law, which is given by:
\[ F = \frac{k_e \cdot |q_1 \cdot q_2|}{r^2} \]
where:
- \( F \) is the electric force,
- \( k_e \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the charges of the particles,
- \( r \) is the distance between the charges.
Given:
- The charge of the iron nucleus, \( q = +26e \),
- The charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \),
- The distance, \( r = 1.5 \times 10^{-12} \, \text{m} \).
Substituting the values:
- \( q_1 = +26e = 26 \times 1.6 \times 10^{-19} \, \text{C} \)
- \( q_2 = -e = -1.6 \times 10^{-19} \, \text{C} \)
Therefore:
\[ F = \frac{8.99 \times 10^9 \cdot |(26 \cdot 1.6 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})|}{(1.5 \times 10^{-12})^2} \]
Calculating inside the absolute value:
\[ |(26 \cdot 1.6 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})| = |(41.6 \times 10^{-19}) \cdot (-1.6 \times 10
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