Answer
$\Delta x=0.44mm$
Work Step by Step
We know that:
$F=\frac{\sigma}{\epsilon_{\circ}}e$
We plug in the known values to obtain:
$F=\frac{2.0\times 10^{-6}}{8.85\times 10^{-12}}\times 1.6\times 10^{-19}=3.6\times 10^{-14}N$
Now;
$W=F.\Delta x$
This can be rearranged as:
$\Delta x=\frac{W}{F}$
We plug in the known values to obtain:
$\Delta x=\frac{1.6\times 10^{-17}}{3.6\times 10^{-14}}=4.4\times 10^{-4}=0.44\times 10^{-3}=0.44mm$
