To find the electric potential energy between two point charges, we use the formula:
[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]\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex]),
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges, and
- [tex]\( r \)[/tex] is the distance between the two charges.
In this problem, we are given:
- [tex]\( q_1 = -4.33 \times 10^{-6} \, \text{C} \)[/tex],
- [tex]\( q_2 = -7.81 \times 10^{-4} \, \text{C} \)[/tex],
- [tex]\( r = 0.525 \, \text{m} \)[/tex].
Substituting these values into the formula, we get:
[tex]\[ U = 8.99 \times 10^9 \frac{(-4.33 \times 10^{-6})(-7.81 \times 10^{-4})}{0.525} \][/tex]
Simplifying inside the fraction first:
[tex]\[ q_1 q_2 = (-4.33 \times 10^{-6}) \times (-7.81 \times 10^{-4}) =
33.7973 \times 10^{-10} = 3.37973 \times 10^{-9} \, \text{C}^2 \][/tex]
(The product of two negative charges is positive.)
Now, divide this product by the distance:
[tex]\[ \frac{3.37973 \times 10^{-9}}{0.525} = 6.43853333333 \times 10^{-9} \][/tex]
Finally, multiply by Coulomb’s constant:
[tex]\[ U = 8.99 \times 10^9 \times 6.43853333333 \times 10^{-9} \approx 57.908100380952376 \, \text{J} \][/tex]
Thus, the electric potential energy between the charges is approximately:
[tex]\[ U \approx 57.91 \, \text{J} \][/tex]
Since the two charges are both negative, their product is positive, and thus the electric potential energy is positive. Therefore, the final answer is:
[tex]\[ U = 57.91 \, \text{J} \][/tex]
