Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To find the force of interaction between the two charged balloons, we use Coulomb's Law. The force [tex]\( F \)[/tex] between two point charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:
[tex]\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
Where:
- [tex]\( k \)[/tex] is Coulomb's constant [tex]\( (9.0 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2) \)[/tex]
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges (both [tex]\(+2.30 \times 10^{-10} \, \text{C}) \)[/tex]
- [tex]\( r \)[/tex] is the separation distance between the charges ( [tex]\(3.20 \times 10^{-1} \, \text{m}) \)[/tex]
Let's input the given values into the formula step by step:
1. Identify the charges:
[tex]\[ q_1 = 2.30 \times 10^{-10} \, \text{C} \][/tex]
[tex]\[ q_2 = 2.30 \times 10^{-10} \, \text{C} \][/tex]
2. Identify the separation distance:
[tex]\[ r = 3.20 \times 10^{-1} \, \text{m} \][/tex]
3. Identify Coulomb's constant:
[tex]\[ k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \][/tex]
4. Plug in these values into Coulomb's law:
[tex]\[ F = 9.0 \times 10^9 \times \frac{(2.30 \times 10^{-10}) \times (2.30 \times 10^{-10})}{(3.20 \times 10^{-1})^2} \][/tex]
Performing the calculations:
- Calculate the product of the charges:
[tex]\[ 2.30 \times 10^{-10} \times 2.30 \times 10^{-10} = 5.29 \times 10^{-20} \][/tex]
- Calculate the square of the distance:
[tex]\[ (3.20 \times 10^{-1})^2 = 1.024 \times 10^{-1} \][/tex]
- Compute the force:
[tex]\[ F = 9.0 \times 10^9 \times \frac{5.29 \times 10^{-20}}{1.024 \times 10^{-1}} \][/tex]
Thus:
[tex]\[ F \approx 9.0 \times 10^9 \times 5.168 \times 10^{-19} = 4.6494140625 \times 10^{-9} \, \text{N} \][/tex]
So, the force of interaction between the two balloons is approximately:
[tex]\[ \boxed{4.6494140625 \times 10^{-9} \, \text{N}} \][/tex]
[tex]\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
Where:
- [tex]\( k \)[/tex] is Coulomb's constant [tex]\( (9.0 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2) \)[/tex]
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges (both [tex]\(+2.30 \times 10^{-10} \, \text{C}) \)[/tex]
- [tex]\( r \)[/tex] is the separation distance between the charges ( [tex]\(3.20 \times 10^{-1} \, \text{m}) \)[/tex]
Let's input the given values into the formula step by step:
1. Identify the charges:
[tex]\[ q_1 = 2.30 \times 10^{-10} \, \text{C} \][/tex]
[tex]\[ q_2 = 2.30 \times 10^{-10} \, \text{C} \][/tex]
2. Identify the separation distance:
[tex]\[ r = 3.20 \times 10^{-1} \, \text{m} \][/tex]
3. Identify Coulomb's constant:
[tex]\[ k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \][/tex]
4. Plug in these values into Coulomb's law:
[tex]\[ F = 9.0 \times 10^9 \times \frac{(2.30 \times 10^{-10}) \times (2.30 \times 10^{-10})}{(3.20 \times 10^{-1})^2} \][/tex]
Performing the calculations:
- Calculate the product of the charges:
[tex]\[ 2.30 \times 10^{-10} \times 2.30 \times 10^{-10} = 5.29 \times 10^{-20} \][/tex]
- Calculate the square of the distance:
[tex]\[ (3.20 \times 10^{-1})^2 = 1.024 \times 10^{-1} \][/tex]
- Compute the force:
[tex]\[ F = 9.0 \times 10^9 \times \frac{5.29 \times 10^{-20}}{1.024 \times 10^{-1}} \][/tex]
Thus:
[tex]\[ F \approx 9.0 \times 10^9 \times 5.168 \times 10^{-19} = 4.6494140625 \times 10^{-9} \, \text{N} \][/tex]
So, the force of interaction between the two balloons is approximately:
[tex]\[ \boxed{4.6494140625 \times 10^{-9} \, \text{N}} \][/tex]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.