Answer:
[tex]U=0.142J[/tex]
Explanation:
The electrostatic potential energy for a pair of charge is given by:
[tex]U=\frac{1}{4\pi E_{o}}\frac{q_{1}q_{2}}{r}[/tex]
Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs of charges.For three equal charges on the corner of an equilateral triangle,the electrostatic energy given by:
[tex]U=\frac{1}{4\pi E_{o}}\frac{q^2}{r}+\frac{1}{4\pi E_{o}}\frac{q^2}{r}+\frac{1}{4\pi E_{o}}\frac{q^2}{r}\\ U=3\frac{1}{4\pi E_{o}}\frac{q^2}{r}\\[/tex]
Substitute the values as q=1.45μC and r=0.400m
So
[tex]U=3\frac{1}{4\pi E_{o}}\frac{q^2}{r}\\ U=3*(9.0*10^9N.m^2/C^2)(\frac{(1.45*10^{-6}C)^2}{0.400m} )\\U=0.142J[/tex]
