Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
To determine how much the electric potential energy has changed, we need to calculate the initial and final electric potential energies and then find the difference between them. Let's go through this step by step:
### Step 1: Define the Given Values
- Charge 1 ([tex]\( q_1 \)[/tex]): [tex]\( 2.55 \times 10^{-4} \)[/tex] Coulombs (C)
- Charge 2 ([tex]\( q_2 \)[/tex]): [tex]\( 7.62 \times 10^{-6} \)[/tex] Coulombs (C)
- Initial distance ([tex]\( r_{\text{initial}} \)[/tex]): [tex]\( 0.350 \)[/tex] meters (m)
- Final distance ([tex]\( r_{\text{final}} \)[/tex]): [tex]\( 1.55 \)[/tex] meters (m)
- Coulomb's constant ([tex]\( k \)[/tex]): [tex]\( 8.99 \times 10^9 \)[/tex] Newton square meters per square Coulomb (N·m²/C²)
### Step 2: Calculate the Initial Electric Potential Energy ([tex]\( U_{\text{initial}} \)[/tex])
The electric potential energy between two point charges is given by the formula:
[tex]\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \][/tex]
For the initial distance ([tex]\( r_{\text{initial}} = 0.350 \)[/tex] meters):
[tex]\[ U_{\text{initial}} = \frac{8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \times 2.55 \times 10^{-4} \, \text{C} \times 7.62 \times 10^{-6} \, \text{C}}{0.350 \, \text{m}} \][/tex]
### Step 3: Calculate the Final Electric Potential Energy ([tex]\( U_{\text{final}} \)[/tex])
For the final distance ([tex]\( r_{\text{final}} = 1.55 \)[/tex] meters):
[tex]\[ U_{\text{final}} = \frac{8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \times 2.55 \times 10^{-4} \, \text{C} \times 7.62 \times 10^{-6} \, \text{C}}{1.55 \, \text{m}} \][/tex]
### Step 4: Find the Change in Electric Potential Energy ([tex]\( \Delta U \)[/tex])
[tex]\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \][/tex]
Using the given numerical results:
[tex]\[ U_{\text{initial}} \approx 49.90991142857143 \, \text{Joules (J)} \][/tex]
[tex]\[ U_{\text{final}} \approx 11.269979999999999 \, \text{Joules (J)} \][/tex]
The change in electric potential energy:
[tex]\[ \Delta U = 11.269979999999999 \, \text{J} - 49.90991142857143 \, \text{J} \][/tex]
[tex]\[ \Delta U \approx -38.63993142857143 \, \text{J} \][/tex]
### Conclusion
The electric potential energy has decreased by approximately [tex]\( 38.64 \)[/tex] Joules when the distance between the charges increased from [tex]\( 0.350 \)[/tex] meters to [tex]\( 1.55 \)[/tex] meters.
### Step 1: Define the Given Values
- Charge 1 ([tex]\( q_1 \)[/tex]): [tex]\( 2.55 \times 10^{-4} \)[/tex] Coulombs (C)
- Charge 2 ([tex]\( q_2 \)[/tex]): [tex]\( 7.62 \times 10^{-6} \)[/tex] Coulombs (C)
- Initial distance ([tex]\( r_{\text{initial}} \)[/tex]): [tex]\( 0.350 \)[/tex] meters (m)
- Final distance ([tex]\( r_{\text{final}} \)[/tex]): [tex]\( 1.55 \)[/tex] meters (m)
- Coulomb's constant ([tex]\( k \)[/tex]): [tex]\( 8.99 \times 10^9 \)[/tex] Newton square meters per square Coulomb (N·m²/C²)
### Step 2: Calculate the Initial Electric Potential Energy ([tex]\( U_{\text{initial}} \)[/tex])
The electric potential energy between two point charges is given by the formula:
[tex]\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \][/tex]
For the initial distance ([tex]\( r_{\text{initial}} = 0.350 \)[/tex] meters):
[tex]\[ U_{\text{initial}} = \frac{8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \times 2.55 \times 10^{-4} \, \text{C} \times 7.62 \times 10^{-6} \, \text{C}}{0.350 \, \text{m}} \][/tex]
### Step 3: Calculate the Final Electric Potential Energy ([tex]\( U_{\text{final}} \)[/tex])
For the final distance ([tex]\( r_{\text{final}} = 1.55 \)[/tex] meters):
[tex]\[ U_{\text{final}} = \frac{8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \times 2.55 \times 10^{-4} \, \text{C} \times 7.62 \times 10^{-6} \, \text{C}}{1.55 \, \text{m}} \][/tex]
### Step 4: Find the Change in Electric Potential Energy ([tex]\( \Delta U \)[/tex])
[tex]\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \][/tex]
Using the given numerical results:
[tex]\[ U_{\text{initial}} \approx 49.90991142857143 \, \text{Joules (J)} \][/tex]
[tex]\[ U_{\text{final}} \approx 11.269979999999999 \, \text{Joules (J)} \][/tex]
The change in electric potential energy:
[tex]\[ \Delta U = 11.269979999999999 \, \text{J} - 49.90991142857143 \, \text{J} \][/tex]
[tex]\[ \Delta U \approx -38.63993142857143 \, \text{J} \][/tex]
### Conclusion
The electric potential energy has decreased by approximately [tex]\( 38.64 \)[/tex] Joules when the distance between the charges increased from [tex]\( 0.350 \)[/tex] meters to [tex]\( 1.55 \)[/tex] meters.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
