Question
A magnesium ion, Mg?*, with a charge of 3.2 × 10-19 C and an oxide ion, O?-, with a charge of -3.2 x 10-19 C, are separated by a distance of 0.45 mm. How.
Answers
GPT-4o mini
To solve this problem, we need to determine the electrical force between the magnesium ion (\( \text{Mg}^{2+} \)) and the oxide ion (\( \text{O}^{2-} \)). We can use Coulomb’s law, which states that the electric force (\( F \)) between two charged particles is given by the formula:
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
where:
- \( F \) is the magnitude of the force between the charges,
- \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the values of the charges,
- \( r \) is the distance between the charges.
In this case:
- The charge of the magnesium ion \( q_1 = 3.2 \times 10^{-19} \, C \) (assuming it’s actually \( \text{Mg}^{2+} \)),
- The charge of the oxide ion \( q_2 = -3.2 \times 10^{-19} \, C \),
- The distance \( r = 0.45 \, mm = 0.45 \times 10^{-3} \, m = 4.5 \times 10^{-4} \, m \).
Now we can substitute the values into Coulomb's law.
1. Calculate the product of the charges:
\[
|q_1 q_2| = |(3.2 \times 10^{-19})(-3.2 \times 10^{-19})| = 3.2 \times 10^{-19} \times 3.2 \times 10^{-19} = 1.024 \times 10^{-37} \, C^2
\]
2. Substitute the values into the formula for the force:
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
3. Calculate \( r^2 \):
\[
r^2 = (4.5 \times 10^{-4})^2 = 2.025 \times 10^{-7} \, m^2
\]
4. Now substitute everything into the formula:
\[
F = (8.99 \times 10^9) \frac{1.024 \times 10^{-37}}{2.025 \times 10^{-7}}
\]
Calculating this gives:
\[
F \approx (8.99 \times 10^9) \times (5.05 \times 10^{-31}) \approx 4.54 \times 10^{-21} \, N
\]
Thus, the magnitude of the force between the magnesium ion and the oxide ion is approximately \( 4.54 \times 10^{-21} \, N \).
Please clarify if you meant to provide specific details regarding other attributes, such as acceleration or potential energy, for further guidance!
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
where:
- \( F \) is the magnitude of the force between the charges,
- \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the values of the charges,
- \( r \) is the distance between the charges.
In this case:
- The charge of the magnesium ion \( q_1 = 3.2 \times 10^{-19} \, C \) (assuming it’s actually \( \text{Mg}^{2+} \)),
- The charge of the oxide ion \( q_2 = -3.2 \times 10^{-19} \, C \),
- The distance \( r = 0.45 \, mm = 0.45 \times 10^{-3} \, m = 4.5 \times 10^{-4} \, m \).
Now we can substitute the values into Coulomb's law.
1. Calculate the product of the charges:
\[
|q_1 q_2| = |(3.2 \times 10^{-19})(-3.2 \times 10^{-19})| = 3.2 \times 10^{-19} \times 3.2 \times 10^{-19} = 1.024 \times 10^{-37} \, C^2
\]
2. Substitute the values into the formula for the force:
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
3. Calculate \( r^2 \):
\[
r^2 = (4.5 \times 10^{-4})^2 = 2.025 \times 10^{-7} \, m^2
\]
4. Now substitute everything into the formula:
\[
F = (8.99 \times 10^9) \frac{1.024 \times 10^{-37}}{2.025 \times 10^{-7}}
\]
Calculating this gives:
\[
F \approx (8.99 \times 10^9) \times (5.05 \times 10^{-31}) \approx 4.54 \times 10^{-21} \, N
\]
Thus, the magnitude of the force between the magnesium ion and the oxide ion is approximately \( 4.54 \times 10^{-21} \, N \).
Please clarify if you meant to provide specific details regarding other attributes, such as acceleration or potential energy, for further guidance!