Answer:
0.12959085 J
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
q = Charge = 1.55 μC
d = Distance between charge = 0.5 m
Electric potential energy is given by
[tex]U=k\dfrac{q^2}{d}[/tex]
In this system with three charges which are equidistant from each other
[tex]U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}[/tex]
[tex]\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J[/tex]
The potential energy of the system is 0.12959085 J
