Respuesta :
Explanation:
The given magnitude of three point charges is as follows.
[tex]q_{1} = q_{2} = q_{3} = q[/tex]
where, q = [tex]1.30 \times 10^{-6} C[/tex] (as [tex]1 \mu C = 10^{-6} C[/tex])
Length of the side is represented as d = 0.4 m
Let U be the electric potential energy of the system and it is represented as follows.
U = [tex]U_{12} + U_{13} + U_{23}[/tex]
= [tex]\frac{1}{4 \pi \epsilon_{o}} \frac{3q^{2}}{d}[/tex]
= [tex]\frac{8.99 \times 10^{9} Nm^{2}/C^{2} \times 3 \times (1.30 \times 10^{-6})^{2}}{0.4}[/tex]
= [tex]11.39 \times 10^{-2}[/tex] J
Thus, we can conclude that the electric potential energy U of the given system is [tex]11.39 \times 10^{-2}[/tex] J.
