# Homework Help: Electrostatic Forcce

1. Sep 3, 2010

### popo902

Electrostatic Forcce :(

1. The problem statement, all variables and given/known data
In Fig. 21-26, particle 1 of charge q1 = +1.4 µC and particle 2 of charge q2 = -4.2 µC, are held at separation L = 12 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) x and (b) y coordinates of particle 3?

2. Relevant equations
F = (Q1 * Q2)K/ r^2

3. The attempt at a solution

At first i thought i could go about this problem setting the particle to the left of the two
i figured that the charge on Q2 > Q1 so its pull would be compensated by Q1's push
(i made my particle three positive)
then i set F31 = F32 so that their forces would balance out to 0
that would make my y placemnt 0 as well
but now...im stuck
i did som canceling until i finally go to this:

(r^2) Q2
------------ = ---------
(0.12 +r)^2 Q1

or
(r^2) / (0.12 +r)^2 = Q2/ Q1

and i don't know wehre to go from there
so...
is there a way out or was i supposed to envision particle three somewhere else??

Last edited: Sep 3, 2010
2. Sep 3, 2010

### gerben

Re: Electrostatic Forcce :(

If you put your particle on left or right side of q1 and q2 so that it is closest to the one that pulls, you can set up two equations:

Fpull = q3*1.4/d^2
Fpush = q3*4.2/(d+12)^2

when the push and pull forces are equal...

3. Sep 3, 2010

### popo902

Re: Electrostatic Forcce :(

Thats wuht i did
and i simplified it to get this

(r^2) / (0.12 +r)^2 = Q2/ Q1

i dint plug in the chrge values for Q1 and 2 yet
but i hope u see the jist
i pretty much started with waht u gave me, (except ur d was my r of course)
but from the equation above i can't figure out how to solve for r
so i was wondering if there was any other way to solve it?? :S

4. Sep 3, 2010

### gerben

Re: Electrostatic Forcce :(

just equate them and solve the equation... here is a start:

q3*1.4/r^2 = q3*4.2/(r+12)^2 (--divide both sides by q3)
1.4/r^2 = 4.2/(r+12)^2 (--evaluate right side)
1.4/r^2 = 4.2/(r^2+24r+144) (--multiply both side by (r^2)(r^2+24r+144) )
1.4(r^2+24r+144) = 4.2r^2
etc....

Last edited: Sep 3, 2010