Answer
The work required is $~~-1.93~J$
Work Step by Step
We can find the distance between the two points:
$r = \sqrt{(\Delta x)^2+(\Delta y)^2}$
$r = \sqrt{(5.50~cm)^2+(1.00~cm)^2}$
$r = 5.59~cm$
We can find the electric potential energy:
$U = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{r}$
$U = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~\frac{(3.00\times 10^{-6}~C)(-4.00\times 10^{-6}~C)}{0.0559~m}$
$U = -1.93~J$
The work required is equal to the electric potential energy of the system.
The work required is $~~-1.93~J$.
