Answer:
Distance between two point charges, r = 0.336 meters
Explanation:
Given that,
Charge 1, [tex]q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=2.4\ \mu C=2.4\times 10^{-6}\ C[/tex]
Electric potential energy, U = -0.5 J
The electric potential energy at a point r is given by :
[tex]U=k\dfrac{q_1q_2}{r}[/tex]
[tex]r=k\dfrac{q_1q_2}{U}[/tex]
[tex]r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}[/tex]
r = 0.336 meters
So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.
