Coulombs Law with one unknown charge

AI Thread Summary
The discussion revolves around calculating the unknown charge q3 using Coulomb's Law, given the coordinates and values of charges q1 and q2. The force on q1 is directed in the -x direction, leading to the conclusion that the y-components of the forces from q2 and q3 must cancel each other out. Participants suggest that assuming q3 equals q2 simplifies the problem significantly. By setting the sum of the y-components of the forces to zero, the relationship between the charges is clarified. The discussion concludes with a sense of resolution and appreciation for the collaborative problem-solving process.
FrogPad
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y(m)
/\
2 | q3
|
|
1 q1
|
|
------q2--> x(m)
2
*graphic* is kind of distorted.
So the coordinates of the charges are:

q1 = (0,1)
q2 = (2,0)
q3 = (2,2)

given:
q1 = 2.5 x 10^-5 C
q2 = 2.0 x 10^-5 C
q3 = ?

Problem:
If the force on q1 points in the -x direction
(a) what is q3.
(b) what is the magnitude of the force on q1.

Notation Key:
Why? Sorry, but the LaTeX seems to be turned off or something.
So I hope this is ok.

r_ab = distance from a to b
R_ab = unit vector from a to b

Coulombs law:
F_ab=k q_a q_b
--------- R
C

r_13 = sqrt(1+2^2) = sqrt(5)m
r_23 = 2m

R_23 = -J = <0,-1>
R_13 = (-2I-J)/sqrt(5) = <(-2sqrt(5))/5,-sqrt(5)/5>

F_net_3 = F_23 + F_13 from superposition principle

F_23 = (9 x 10^9 N m^2) (2.0 x 10^-5 C) q3 C R_23
---------------- ---------------
C^2 (2m)^2

F_13 = (9 x 10^9 N m^2) (2.5 x 10^-5 C) q3 C R_23
---------------- ---------------
C^2 (2m)^2

F_23 = <0, -1 * (1.8 x 10^5 q3)/5> N
F_13 = <(2.25 x 10^5 q3)/5 (-2)sqrt(5)/5
,2.25 x 10^5 q3)/5 (sqrt(5))/5>

F_23 = <0, -3.6 x 10^4 q3> N
F_13 = <-40249.2 q3, -20124.6 q3> N


Now this is where I get stuck since I have to many unknowns.
I just don't know what to do from here... ANY help would be
amazing. Thanks.
 
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your r_13 and r_23 is completely wrong already, I didn't check the rest...
do you having difficulty finding distance?
 
Sorry about that. Actually the mistake was in the coordinate system. q3 is supposed to be (2,2) not (0,2) like I had... so I guess I have trouble typing in proper numbers, not finding the distance :)

P.S. It looks god awefull without latex. Is it turned off or something?
 
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draw the graph, use symmetry argue q3 is equal to q2(hopefully you can see that), it will save you a lot of time
then find the x component of the force (should be easy for you)
 
I don't really see why q3 is equal to q2. But, I'll just go under the assumption that it is and work though the process.
Thank you for the help :)
 
FrogPad said:
I don't really see why q3 is equal to q2. But, I'll just go under the assumption that it is and work though the process.
Thank you for the help :)

The y-components of the two forces have to cancel, since the force at q1 is in the -x direction.

Write an equation for the sum of the 2 y-components and set it equal to zero. Then you'll see that q3=q2.
 
Thank you guys so much. I guess I was just making the problem harder then it was. I definitely see why the forces cancel if q3=q2. Ok cool... :) Just glad I got that done.
By the way, this board is awesome.
 
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