Answer
$W=2.30\times 10^{-28}J$
Work Step by Step
We can find the electric potential as
$V=\frac{K(2e)}{2D}+\frac{K(e)}{D}$
We plug in the known values to obtain:
$V=\frac{9\times 10^9\times (2\times1.6\times 10^{-19})}{2(4)}+\frac{9\times 10^9\times (1.6\times 10^{-19})}{4}=7.2\times 10^{-10}V$
Now we can find the work done as
$W=q\Delta V$
We plug in the known values to obtain:
$W=2\times 1.6\times 10^{-19}\times 7.2\times 10^{-10}=2.30\times 10^{-28}J$
