Answer :
To calculate the electric potential energy between two point charges, we use the formula derived from Coulomb's law:
[tex]\[ U = k \frac{q_1 q_2}{r} \][/tex]
where:
- [tex]\( U \)[/tex] is the electric potential energy,
- [tex]\( k \)[/tex] is Coulomb's constant [tex]\((8.9875517873681764 \times 10^9 \, \text{N m}^2/\text{C}^2)\)[/tex],
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the electric charges,
- [tex]\( r \)[/tex] is the distance between the charges.
Given:
- [tex]\( q_1 = 0.00287 \, C \)[/tex],
- [tex]\( q_2 = -0.00555 \, C \)[/tex],
- [tex]\( r = 6.52 \, m \)[/tex].
The charges must be multiplied together and then divided by the distance. The sign of the product of the charges will affect the sign of the electric potential energy.
Let's calculate:
1. Multiply the two charges:
[tex]\[ q_1 \cdot q_2 = 0.00287 \, C \times -0.00555 \, C = -1.59285 \times 10^{-5} \, C^2 \][/tex]
2. Multiply this result by Coulomb's constant [tex]\( k \)[/tex]:
[tex]\[ k \cdot (q_1 \cdot q_2) = 8.9875517873681764 \times 10^9 \, \text{N m}^2/\text{C}^2 \times -1.59285 \times 10^{-5} \, C^2 = -1.4327462706565748 \times 10^5 \, \text{N m}^2/\text{C} \][/tex]
3. Divide by the distance [tex]\( r \)[/tex]:
[tex]\[ U = \frac{-1.4327462706565748 \times 10^5 \, \text{N m}^2/\text{C}}{6.52 \, m} \approx -2.195678200078288 \times 10^4 \, J \][/tex]
[tex]\[ U \approx -21956.782000781288 \, J \][/tex]
So, the electric potential energy between the two charges is approximately [tex]\( -21956.782 \, J \)[/tex].
[tex]\[ U = k \frac{q_1 q_2}{r} \][/tex]
where:
- [tex]\( U \)[/tex] is the electric potential energy,
- [tex]\( k \)[/tex] is Coulomb's constant [tex]\((8.9875517873681764 \times 10^9 \, \text{N m}^2/\text{C}^2)\)[/tex],
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the electric charges,
- [tex]\( r \)[/tex] is the distance between the charges.
Given:
- [tex]\( q_1 = 0.00287 \, C \)[/tex],
- [tex]\( q_2 = -0.00555 \, C \)[/tex],
- [tex]\( r = 6.52 \, m \)[/tex].
The charges must be multiplied together and then divided by the distance. The sign of the product of the charges will affect the sign of the electric potential energy.
Let's calculate:
1. Multiply the two charges:
[tex]\[ q_1 \cdot q_2 = 0.00287 \, C \times -0.00555 \, C = -1.59285 \times 10^{-5} \, C^2 \][/tex]
2. Multiply this result by Coulomb's constant [tex]\( k \)[/tex]:
[tex]\[ k \cdot (q_1 \cdot q_2) = 8.9875517873681764 \times 10^9 \, \text{N m}^2/\text{C}^2 \times -1.59285 \times 10^{-5} \, C^2 = -1.4327462706565748 \times 10^5 \, \text{N m}^2/\text{C} \][/tex]
3. Divide by the distance [tex]\( r \)[/tex]:
[tex]\[ U = \frac{-1.4327462706565748 \times 10^5 \, \text{N m}^2/\text{C}}{6.52 \, m} \approx -2.195678200078288 \times 10^4 \, J \][/tex]
[tex]\[ U \approx -21956.782000781288 \, J \][/tex]
So, the electric potential energy between the two charges is approximately [tex]\( -21956.782 \, J \)[/tex].
