Answer:
U = 0.113 J
Explanation:
Electrostatic potential energy of two charges at some distance is given as
[tex]U = \frac{kq_1q_2}{r}[/tex]
since there are three charges on the vertices of the triangle
so here total potential energy is given as
[tex]U_{net} = 3\frac{kq_1q_2}{r}[/tex]
[tex]U_[net} = 3(\frac{(9\times 10^9)((1.65 \mu C)(1.65 \mu C)}{0.650})[/tex]
[tex]U_{net} = 3(0.038 J)[/tex]
[tex]U_{net} = 0.113 J[/tex]
So electrostatic total energy is 0.113 J