Force equalibrium

  • Thread starter leon1127
  • Start date
  • #1
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Three point charges are arranged along the x-axis. Charge q_1=−4.05 nC is located at x_1 = 0.195 m, and charge q_2 = 2.80 nC is at x_2=−0.340 m A positive point charge q_3 is located at the origin.Where along the x-axis can q_3 be placed and the net force on it be zero, other than the trivial answers of x= \pm \infty?
Thx
 

Answers and Replies

  • #2
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show some work.

how do you think it should be solved?
 
  • #3
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[tex]\frac{4.05* q_3}{(-0.34-x)^2} =\frac{-2.8 * q_3}{(0.195+x)^2} [\tex]

I dont know why it doesnt work
 
Last edited:
  • #4
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may anyone?
 
  • #5
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draw the diagram:

|------x2-----------------------x----------------x1----------|

if x is some value, then to find the distance, you have to use the distances as (x-x2) and (x1-x)
 
  • #6
486
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how do you know if x will land in the interval between [tex] q_1 and q_2 [\tex]
the problem i get is when i assume the q3 will at equalibrium, x become imaginary when i take the square root

2.8/x^2 == -4.05 / (0.535+x)^2

i have shifted the system to the right. Then the negative sign became PAIN under the square root
 
Last edited:
  • #7
217
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crap.

i'm sorry, i didn't read the question carefully enough. as far as i can tell, it has to be either left of q_2 or right of q_1. just from eyeballing the figure, it looks like it will be to the left of q_2 because the smaller repelling charge needs to be closer to cancel out the larger attracting charge.
 
  • #8
486
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nvm i got it.
Thank you everyone
 

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