Answer
$\sum W=6.90\times10^{-18}J$
Work Step by Step
$d=1.0\times10^{-10}m$
$\sum W=3\times k\frac{q_1q_1}{d}$
$=3\times(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(1.60\times10^{-19}C)^2}{1.0\times10^{-10}m}=6.90\times10^{-18}J$
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