Certainly! Let's find the electric potential energy between two charges. We’ll go through the steps methodically.
### Given:
1. The first charge, [tex]\( q_1 = -4.33 \times 10^{-6} \)[/tex] Coulombs.
2. The second charge, [tex]\( q_2 = -7.81 \times 10^{-4} \)[/tex] Coulombs.
3. The distance between the charges, [tex]\( r = 0.525 \)[/tex] meters.
4. Coulomb's constant, [tex]\( k = 8.988 \times 10^9 \)[/tex] [tex]\( \text{N m}^2/\text{C}^2 \)[/tex].
### Formula for Electric Potential Energy:
The electric potential energy ([tex]\( U \)[/tex]) between two point charges is given by:
[tex]\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \][/tex]
### Steps to Calculate:
1. Substitute the values into the formula:
[tex]\[ U = \frac{ (8.988 \times 10^9) \cdot (-4.33 \times 10^{-6}) \cdot (-7.81 \times 10^{-4}) } { 0.525 } \][/tex]
2. Multiplication of like terms:
- Multiply the values within the numerator.
[tex]\[ -4.33 \times 10^{-6} \times -7.81 \times 10^{-4} = 3.380634 \times 10^{-9} \][/tex]
- Note that the product of two negative numbers is positive.
3. Numerator after incorporating Coulomb's constant:
[tex]\[ (8.988 \times 10^9) \times 3.380634 \times 10^{-9} = 30.394713592 \][/tex]
4. Dividing by the distance:
[tex]\[ \frac{30.394713592}{0.525} = 57.8952176 \][/tex]
### Conclusion:
The electric potential energy between these two charges is therefore:
[tex]\[ U = 57.8952176 \text{ Joules} \][/tex]
This result takes into account the appropriate signs and constants, confirming the calculations are precise and consistent.
So, the electric potential energy of the two charges is [tex]\( U = 57.8952176 \text{ Joules} \)[/tex].
