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Where on the x-axis will a third charge have no net force

  • Thread starter bbraves7
  • Start date
1. The problem statement, all variables and given/known data

Where on the x-axis would the net force on a (-2) micro Coulomb charge be zero?

*picture attached*

2. Relevant equations

F = k*(q1q2)/d^2

3. The attempt at a solution

I think the answer is d= 6, making the position on the x-axis = 7. However, when solving my equation, I have to take the square root of a negative number, which obviously has no real answer. I couldn't seem to logic my way around it, so what am I missing? Other than expressing the answer in terms of i, how can I get around this??
 

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SammyS

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9 μC = 9×10‒6 C , Similarly ‒4 μC = ‒4×10‒6 C and ‒2 μC = ‒2×10‒6 C.

You have written them as 9‒6 C, ‒4‒6 C, and ‒2‒6 C .

(Maybe you are simply using a non-standard notation.)
 

SammyS

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You need F1 + F2 = 0.

So that F1 = ‒ F2

That should take care of your signs.
 
9 μC = 9×10‒6 C , Similarly ‒4 μC = ‒4×10‒6 C and ‒2 μC = ‒2×10‒6 C.

You have written them as 9‒6 C, ‒4‒6 C, and ‒2‒6 C .

(Maybe you are simply using a non-standard notation.)
Yeah, I just write it like that as a kind of short hand.

Anyways, I figured out what I was doing wrong. The equation should only be used to calculate the magnitude of the force, so I shouldn't put the negative in the equation for the -4 micro Coulomb charge. That takes care of it. Thanks for the reply though.
 

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