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Coulombs Law with one unknown charge

 
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Jan25-05, 01:48 AM   #1
 

Coulombs Law with one unknown charge


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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Jan25-05, 01:54 AM   #2
 
your r_13 and r_23 is completely wrong already, I didn't check the rest....
do you having difficulty finding distance?
Jan25-05, 12:05 PM   #3
 
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?
Jan25-05, 01:12 PM   #4
 

Coulombs Law with one unknown charge


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)
Jan25-05, 01:34 PM   #5
 
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 :)
Jan25-05, 01:57 PM   #6
 
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Quote by FrogPad
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.
Jan26-05, 12:57 PM   #7
 
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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