Chapter 20 - Electric Potential and Electric Potential Energy - Problems and Conceptual Exercises - Page 719: 38

$-0.27J$

We can find the required work done as follows: $W=\Delta U=0-(\frac{Kq_1q_2}{r_{12}}+\frac{Kq_1q_3}{r_{13}}+\frac{Kq_2q_3}{r_{23}})$ $W=-K(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_{23}}{r_{23}})$ We plug in the known values to obtain: $W=-(8990)[\frac{(-6.1)(2.7)}{0.25}+\frac{(-6.1)(3.3)}{0.16}+\frac{(2.7)(-3.3)}{0.30}]$ $W=-0.27J$