Question
Describe the relationship between the force and the magnitudes of the charges. A:
What happens to the force when the distance is halved? How does this relate to Coulomb’s Law? A:
Look up the mathematical expression for Coulomb’s Law. A:
Predict the force if one charge is +3.0 nC, the other is −2.0 nC, and they are 1.5 meters apart using Coulomb’s Law formula. Show your work. A:
Answer each question in 2-3 sentences
What happens to the force when the distance is halved? How does this relate to Coulomb’s Law? A:
Look up the mathematical expression for Coulomb’s Law. A:
Predict the force if one charge is +3.0 nC, the other is −2.0 nC, and they are 1.5 meters apart using Coulomb’s Law formula. Show your work. A:
Answer each question in 2-3 sentences
Answers
GPT-4o mini
The relationship between the force and the magnitudes of the charges is directly proportional according to Coulomb's Law. This means that if one or both of the charges increase, the force of attraction or repulsion between them also increases.
When the distance between two charges is halved, the force between them increases by a factor of four. This relationship is described by Coulomb’s Law, which states that the force is inversely proportional to the square of the distance between the charges.
Coulomb’s Law is mathematically expressed as \( F = k \frac{|q_1 \cdot q_2|}{r^2} \), where \( F \) is the force between the charges, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.
To predict the force when one charge is +3.0 nC (or \( 3.0 \times 10^{-9} \) C) and the other is -2.0 nC (or \( -2.0 \times 10^{-9} \) C), and they are 1.5 m apart, we can substitute the values into Coulomb's Law. Thus:
\[ F = 8.99 \times 10^9 \frac{|3.0 \times 10^{-9} \cdot (-2.0 \times 10^{-9})|}{(1.5)^2} \]
Calculating this gives:
\[ F = 8.99 \times 10^9 \frac{6.0 \times 10^{-18}}{2.25} \]
\[ F = 8.99 \times 10^9 \cdot 2.67 \times 10^{-18} \approx 2.40 \times 10^{-8} \, \text{N} \]
Thus, the force between the two charges is approximately \( 2.40 \times 10^{-8} \) N, directed towards each other due to the opposite signs of the charges.
When the distance between two charges is halved, the force between them increases by a factor of four. This relationship is described by Coulomb’s Law, which states that the force is inversely proportional to the square of the distance between the charges.
Coulomb’s Law is mathematically expressed as \( F = k \frac{|q_1 \cdot q_2|}{r^2} \), where \( F \) is the force between the charges, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.
To predict the force when one charge is +3.0 nC (or \( 3.0 \times 10^{-9} \) C) and the other is -2.0 nC (or \( -2.0 \times 10^{-9} \) C), and they are 1.5 m apart, we can substitute the values into Coulomb's Law. Thus:
\[ F = 8.99 \times 10^9 \frac{|3.0 \times 10^{-9} \cdot (-2.0 \times 10^{-9})|}{(1.5)^2} \]
Calculating this gives:
\[ F = 8.99 \times 10^9 \frac{6.0 \times 10^{-18}}{2.25} \]
\[ F = 8.99 \times 10^9 \cdot 2.67 \times 10^{-18} \approx 2.40 \times 10^{-8} \, \text{N} \]
Thus, the force between the two charges is approximately \( 2.40 \times 10^{-8} \) N, directed towards each other due to the opposite signs of the charges.