Electrostatics: Finding Equilibrium with Three Charges

[PsychonautQQ]

Homework Statement

Say there is a charge of +3 at the origin and a charge of -7 at .5m Where would a third charge of arbitrary sign have to be for equilibrium to be reached?

Homework Equations

The Attempt at a Solution

so I've widdled this down to 3 / r^2 = -7/(.5+r)^2 but am having problems solving for r. Algebra is tough! anyone want to give me tips here? In the equation I have I already divided out the third charge and the k term.

 

[haruspex]

so I've widdled this down to 3 / r^2 = -7/(.5+r)^2 but am having problems solving for r. Algebra is tough!

It's particularly tough when you try to make a determinedly positive term equal an insistently negative one . The signs depend on whether the test charge is placed left or right of the given charges.

 

[PsychonautQQ]

Ahh right! So I could just do 3/r^2 = 7/(.5+r)^2?
Following up would give...
sqrt(3/7) = (.5+r)/r??
sqrt(3/7)*r = .5 + r
0 = .5 + r - sqrt(3/7)*r
0 = .5 + .345346*r
-.5 / .345346 = r would mean r is -1.4478.. But the back of the book says otherwise ;-( what did I do wrong here?

 

[PsychonautQQ]

Yo shane, if you are reading this you should email me the title/author of your textbook and the problem number and i'll have a clearer solution posted here tonight

 

[PsychonautQQ]

yo check your email shane

 

[haruspex]

Ahh right! So I could just do 3/r^2 = 7/(.5+r)^2?

Right.

Following up would give...
sqrt(3/7) = (.5+r)/r??

Wrong.