Two point charges are located on the $x$ -axis,  $q_1=-1\mu C$ at  $x=0$ and $q_2=+1\mu C$ at $x=1m$.

Find the work that must be done by an external force to bring a third point charge $q_3=+1\mu C$ from infinity to $x=2m$.

The work that must be done on $q_3$ by an external force is equal to the different of potential energy $U$ when the charge is at $x=2m$ and the potential energy when it is at infinity

$\therefore W=U_f-U_i$

$=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{q_3{q}_2}{{\left(r_{32}\right)}_f}+\frac{q_3{q}_1}{{\left(r_{31}\right)}_f}+\frac{q_2{q}_1}{{\left(r_{21}\right)}_f}\right]-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{q_3{q}_2}{{\left(r_{32}\right)}_i}+\frac{q_3{q}_1}{{\left(r_{31}\right)}_i}+\frac{q_2{q}_1}{{\left(r_{21}\right)}_i}\right]$

Here, ${\left(r_{21}\right)}_i={\left(r_{21}\right)}_f$

and ${\left(r_{32}\right)}_i={\left(r_{31}\right)}_i=\infty$

$\therefore W=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{q_3{q}_2}{{\left(r_{32}\right)}_f}+\frac{q_3{q}_1}{{\left(r_{31}\right)}_f}\right]$

Substituting the values, we have

$W=\left(9.0\times {10}^9\right)\left({10}^{-12}\right)\left[\frac{\left(1\right)\left(1\right)}{(1.0)}+\frac{\left(1\right)(-1)}{(2.0)}\right]$ 

$=4.5\times {10}^{-3}J$