Answer:
-898.3 J
Explanation:
The electric potential energy of a system of two charges is given by
[tex]U=\frac{kq_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1, q2 are the two charges
r is the separation between the two charges
In this problem, at the beginning we have:
[tex]q_1=q_2=2.55\cdot 10^{-4}C[/tex] are the two charges
[tex]r_1=0.500 m[/tex] is the initial separation
While after the charges have been released, we have:
[tex]r_2=2.15 m[/tex] is the final distance
Therefore, the change in electric potential energy is:
[tex]\Delta U=U_2 -U_1 = \frac{kq_1 q_2}{r_2}-\frac{kq_1q_2}{r_1}=kq^2(\frac{1}{r_2}-\frac{1}{r_1})=\\=(9\cdot 10^9)(2.55\cdot 10^{-4})^2(\frac{1}{2.15}-\frac{1}{0.500})=-898.3 J[/tex]
