1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
[tex]d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} [/tex]
substituting the coordinates of the two charges, we get
[tex]d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m [/tex]
2) Then, we can calculate the electrostatic force between the two charges [tex]q_1[/tex] and [tex]q_2[/tex], which is given by
[tex]F=k_e \frac{q_1 q_2}{d^2} [/tex]
where [tex]k_e=8.99\cdot10^{9} Nm^2C^{-2}[/tex] is the Coulomb's constant.
Substituting numbers, we get
[tex]F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N[/tex]
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.
