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Coulombs Law with one unknown charge |
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| Jan25-05, 01:48 AM | #1 |
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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. |
| Jan25-05, 01:54 AM | #2 |
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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? |
| Jan25-05, 12:05 PM | #3 |
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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 |
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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 |
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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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Recognitions:
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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 |
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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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