As we know that change in the potential energy of the system must be equal to the work done to bring the system of charges
here we know that
[tex]U = 3\frac{kq_1q_2}{r^2}[/tex]
also here we have
[tex]q_1 = q_2 = 1.6 \times 10^{-19} C[/tex]
[tex]r = 9 \times 10^{-10} C[/tex]
now we will have
[tex]U = \frac{9 \times 10^9 (1.6 \times 10^{-19})^2}{(9\times 10^{-10})^2}[/tex]
[tex]U = 2.84 \times 10^{-10} J[/tex]
So here work done to bring three charges at the vertices of a triangle is given as
[tex]W = 2.84 \times 10^{-10} J[/tex]
