Two fixed charges and one free to move problem.

  • Thread starter Cmon101
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  • #1
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Homework Statement


Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1.
Q1= 1.39x10-6 C and Q2= -3.22 x10-6 C are fixed at their positions, distance 0.301 m apart, and the charge Q3= 3.33 x10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.



Homework Equations


F=(K(Q1Q2))/r^2



The Attempt at a Solution


So I attempted to solve it by using X as the distance from Q1 to Q3 and built this equation:

[k(Q1Q3)]/(X^2)+[k(Q2Q3)]/[(x-.301)^2)=0

Since .301 is the distance from 1 to 2 which is fixed I assumed X-.301 was the correct way to calculate distance from 2 to 3 for that portion of the equation.

When I go throught this I get to the point where I'm at [.0417/(x^2)]= [-.0965/(x-.301)^2].

I then plug this into my graphing calculator (one side as one function, the other as another function) and then excute the intercept function to easily find X, but there is no intercept meaning that my equation above for X is wrong.

Any help is appreciated. Thank you in advance.
 
Last edited:

Answers and Replies

  • #2
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Hi Cmon101
welcome to PF


instead try to write the forces b/w Q1Q2 and Q1Q3
 
  • #3
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How do I use the force of Q1 on Q2 after I find it? I thought since one and two are stationary that force did not matter. Sorry I'm kind of slow at understanding this stuff.
 
  • #4
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I'm sorry i was reading 2 ques. and i mixed them.

Ok.... now so you know that q1, q3 is +ve and q2 is negative and also mag. of q1 < mag of q2 ... can you tell me should q3 be between q1 and q2 or not?
 
  • #5
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OMG good call. I kept thinking it had to either be Q1------Q2----Q3 or Q1---Q3---Q2 and I kept thinking that just doesnt make sense logically cause of the charges but Q3------Q1----Q2 does make sense! Alright altering my equation to take the new positioning into account. Will tell you the result.
 
  • #6
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Damn. Still didn't work. So if we assume the setup is like this
Q3(3.33 uC)------X----Q1(1.39uC)-----.301m-----Q2(-3.22 uC)

The equation would turn into this correct:

K(Q3Q1)/ X2 + K(Q3Q2) /(.301+X)2 = 0

Or more plainly K(Q3Q1)/ X2 = K(Q3Q2) /(.301+X)2 If you use the absolute values of the charges.
 
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  • #7
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Wait I guess it could look like this Q1(1.39uC)-----.301m-----Q2(-3.22 uC)-----Q3(3.33)
X being the distance from 1 to 3.

Actually does this make sense? I think Q3 on the left is actually the right setup cause in this one wouldnt it be impossible for the forces to equilibrate no matter how close or far away charge 3 was on the right.
 
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  • #8
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Damn. Still didn't work. So if we assume the setup is like this
Q3(3.33 uC)------X----Q1(1.39uC)-----.301m-----Q2(-3.22 uC)

The equation would turn into this correct:

K(Q3Q1)/ X2 + K(Q3Q2) /(.301+X)2 = 0
well it has to work this way

and the next one wont work as q2 has more mag. than q1 so in order of zero net force, Q3 should be near Q1

maybe answer given to you is not correct
 
  • #9
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what is your answer?
 
  • #10
11
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So K(Q3Q1)/ X2 + K(Q3Q2) /(.301+X)2 is the same as:

K(Q3Q1)/ X2 = K(Q3Q2) /(.301+X)2 (If we use the absolute value for the charges)
And Then we can turn this into

(.301+X)2 / (X)2 = [K(Q3Q2)]/[K(Q3Q1)]

Then the Ks cancel on the right side leaving you with

(.301+X)2 / (X)2 = (Q3Q2)/(Q3Q1)

Then we square root both sides and it turns into
(.301+X)/X = SQROOT[(Q3Q2)/(Q3Q1)]

Does all of this algebra look correct so far?
 
  • #11
Delta2
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I think your very first equation will be correct if u remove the minus sign from right hand side.
 
  • #12
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Then from (.301+X)/X = SQROOT[(Q3Q2)/(Q3Q1)]

We divide the left side of the equation to get:

.301/X+1=SQROOT[(Q3Q2)/(Q3Q1)]

Move the 1:

.301/X=SQROOT[(Q3Q2)/(Q3Q1)]-1

Move the X:
.301=[SQROOT[(Q3Q2)/(Q3Q1)]-1]X

Get X alone:
(.301)/[SQROOT[(Q3Q2)/(Q3Q1)]-1]=X

And thats the end of it does all of that look right about to plug it in
 
  • #13
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I got .58 meters and still incorrect. I do not have the actual answer.
 
  • #14
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I think your very first equation will be correct if u remove the minus sign from right hand side.
But if we assume 3 must be on the left of one done we have to change x-.301 to x+.301.
 
  • #15
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I forgot to put a negative sign in front of distance since it was to the left. -.58 m is correct. Thanks so much for sticking through this with me and helping me out cupid!
 
  • #16
Delta2
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But if we assume 3 must be on the left of one done we have to change x-.301 to x+.301.
Using the initial equation with x-.301 you ll just find 2 values for x, one negative and one positive. You keep only the negative one cause in the positive solution the forces do not cancel out but addup.
 
  • #17
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Using the initial equation with x-.301 you ll just find 2 values for x, one negative and one positive. You keep only the negative one cause in the positive solution the forces do not cancel out but addup.
Ah got ya
 

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