Finding the Position and Charge for a Zero Resultant Force in Coulomb's Law

[EvanQ]

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.

 

[Shooting Star]

((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

From ((Qx16)/(x^2 -4x+4) = ((Qx9)/(x^2)), you write 16/(-4x+4) = 9?

Cross-multiply and solve properly.

Does the charge have to be -ve?

 

[EvanQ]

alright got it now, thanks a lot for your help.

i was about 80% through typing out my reworking to see if you could check it when i accidentally clicked the back button on my mouse and lost it all, and i really can't put myself through typing it again lol.

 

[rado5]

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?.

I think your answer is wrong. Because q$$_{}2$$=9$$\mu$$c is placed at the origin so q$$_{3}$$ must be placed between q$$_{2}$$ and q$$_{1}$$ and if X is the distance that q$$_{3}$$ is placed from the origin the answer will be " X=86cm ".Because kq$$_{}3$$q$$_{}2$$/x$$^{}2$$=kq$$_{}3$$q$$_{}1$$/(2-X)$$^{}2$$ and of course the charge must be minus

 

[Shooting Star]

... and of course the charge must be minus

You sound very confident. Think about it for a while.