Answer:
[tex]W_{total}=4.5*10^{-8}J[/tex]
Explanation:
Remember that electric potential can be written as:
[tex]V=\frac{kQ}{r}[/tex],
Where V is the electric potential, k is Coulomb's constant, Q is a point charge, and r is the distance from the point charge. Also, we can write the electric potential as:
[tex]V=\frac{W}{q}[/tex],
where W is the work made to move a charge from infinitely far apart to a certain distance, and q the point charge were are moving.
From all this we can get an expression for the work:
[tex]W=\frac{kQq}{r}[/tex]
We are going to let
[tex]q_{1}=-1.00nC\\q_{2}=2.00nC\\q_{3}=3.00nC[/tex]
To take the first charge [tex]q_{1}[/tex] from infinitely far apart to one of the vertices of the triangle, since there is no electric field and charges, we make no work.
Next, we will move [tex]q_{2}[/tex] . Now, [tex]q_{1}[/tex] is a vertice of the triangle, and we want [tex]q_{2}[/tex] to be 20.cm apart from [tex]q_{1 }[/tex] so the work we need to do is
[tex]W_{12}=\frac{kq_{1}q_{2}}{(0.20)}\\\\W_{12}=\frac{(9*10^{9})(-1*10^{-9})(2*10^{-9})}{0.20}\\\\W_{12}=-9*10^{-8}J[/tex]
Now, we move the last point charge. Here, we need to take in account the electric potential due to [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. So
[tex]W=W_{13}+W_{23}\\\\\\W=\frac{kq_{1}q_{3}}{0.20}+\frac{kq_{2}q_{3}}{0.20}\\\\W=q_{3}k(\frac{q_{1}}{0.20}+\frac{q_{2}}{0.20})\\\\W=(3*10^{-9})(9*10^{9})(\frac{-1*10^{-9}}{0.20}+\frac{2*10^{-9}}{0.20})\\\\[/tex]
[tex]W=1.35*10^{-7}J[/tex]
Now, the only thing left to do is to find the total work, this can be easily done by adding [tex]W_{12}[/tex] and [tex]W[/tex]:
[tex]W_{total}=W_{12}+W\\W_{total}=1.35*10^{-7}-9*10^{-8}\\W_{total}=4.5*10^{-8}J[/tex]
