Where Should a +2.5Q Charge Be Placed to Nullify the Force?

[K.J.Healey]

Jsut take what you have, and split it up:
$$( 0,222 \vec{i} + 0,250\vec{j}) = \frac{2,5}{ x^2 + y^2} x \vec{i} + y \vec{j}$$

becomes
$$( 0,222 \vec{i}) = \frac{2,5}{ x^2 + y^2} x \vec{i}$$
and
$$( 0,250\vec{j}) = \frac{2,5}{ x^2 + y^2} y \vec{j}$$

Then its basically 2 equations with 2 unknowns.

I would then say solve the top one for y^2 and then plug it into the bottom. Its a lot of algebra from there but it would work. I honestly think splitting up vectors into the two parts gets rid of the point of even having vectors. But it DOES make things easier to computer in classes with dynamics and forces acting on objects. Sum the forces in each direction, set equal to m*a, youll see that a lot.

But you should always make sure you feel comfortable using something like full vector notation before you toss it aside for an easier way.

 

[mad]

When I solve for x and y , I get x= (+-) 3,16 and y= 1,27 ...
Anyway, I already understood your first method, it's a lot more easier.. but I don't know why it never works like you have written.
Thanks a lot for all your time and help, Healey01 :)