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EvanQ
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Homework Statement
Three point charges lie along the x axis. A positive charge q1=16.0mC is at x=2.00m. Another positive charge q2=9.00mC is at the origin.
Where should we put a third charge so the resultant force acting on it is zero?
What should be the sign of the charge?
Homework Equations
F=k(Qq/(r^2))
k=9x10^9 (Nm^2)/(C^2)
The Attempt at a Solution
First of all, the charge must be -ve if there is going to be a net force of zero acting on it between the two positive charges.
Let the distance between the origin charge and the negative charge be x. Therefore the distance between q2 and the negative charge will be 2m - x.
The force between charges 1 and 3, and the force between charges 2 and 3 must be equal if there is to be no net charge, thus F13 = F23.
F13 = 9x10^9 ((Qx9uc)/(x^2))
F23 = 9x10^9 ((Qx16uc)/((2-x)^2))
= 9x10^9 ((Qx16uc)/(x^2 -4x+4)
((Qx16)/(x^2 -4x+4) = ((Qx9)/(x^2))
let Q = 1 for the sake of ratio's.
16/(-4x+4) = 9
4/(-x+1)=9
-x+1 = 0.44
x=0.555556
Therefore the 3rd charge should be places 55.5cm from the origin.
I feel like this is definitely wrong and my algebra is whacked up somewhere... can anyone help showing me where and how to rectify it? thanks.