Coulomb's law-superposition principle

[loreben]

Hi SDNers,
I tried solving the question below but i seem to be getting no where. I guess my main problem with this question is knowing where to place the positive charge /q3/ and why it is placed in that particular position and not elsewhere. Cheers 🙂

Exercise

Three charges lie along the x-axis. The positive charge q1=10micro coulombs is at x= 1m, and the negative charge q2= -2micro coulombs is at the origin. Where must a positive charge q3 be placed on the x-axis so that the resultant force on it is zero?

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[Rabolisk]

I guess my main problem with this question is knowing where to place the positive charge /q3/ and why it is placed in that particular position and not elsewhere.

Three charges lie along the x-axis. The positive charge q1=10micro coulombs is at x= 1m, and the negative charge q2= -2micro coulombs is at the origin. Where must a positive charge q3 be placed on the x-axis so that the resultant force on it is zero?

Those numbers make this a rather ugly problem to solve, but qualitatively it's pretty clear which side of the origin on the x-axis the positive charge has to be placed. Can you tell me which side?

 

[loreben]

I figured it should be on the negative side of the x-axis but am not totally sure why that should be the case.

 

[GreenRabbit]

You use the formula F = kqq /r^2 for the interaction of q3 with both charges.

So F1 = kQ(10uC) / (x- 1m)^2
F2 = kQ(-2uC) / (x- 0m)^2

You want the total force to be 0, so F1 + F2 = 0, therefore you have the equation

kQ(10uC) / (x- 1m)^2 = kQ(2uC) / (x- 0m)^2

Note that X must be negative for this to be true.
k and Q cancel out, leaving you with

10 / (x- 1)^2 = 2/ x^2

5 = (x-1)^2 / x^2 = [(x-1)/x]^2
sqrt(5) = 1 - 1/x
x = 1 / [1 - sqrt(5)]