Can a Third Sphere with Opposite Charge Achieve Zero Net Force?

[xxabr]

Homework Statement

Two spheres, with charges 1.6x10^-5C and 6.4x10^-5C are 2m apart. (The charges have the same sign). In the middle of the two spheres, should a third sphere,, of opposite charge 3.0x10^-6C, be placed if the third sphere has zero net electrical force?

(1.6x10^-5) -------- x -------- (6.4x10^-5)
|_________________2m ________________|

Homework Equations

F1=F2

E=kq/r^2
Fe= kq1q2/r^2
E=Fe/r^2

The Attempt at a Solution

My teacher said we have to use the quadratic formula but I don't know to how to form that.
If you could help me with a starting point that would be great.

F1= x
F2= 2-x

Please help, thanks.

 

[tiny-tim]

welcome to pf!

hi xxabr! welcome to pf!

(try using the X² and X₂ icons just above the Reply box )

My teacher said we have to use the quadratic formula but I don't know to how to form that.
If you could help me with a starting point that would be great.

F1= x
F2= 2-x

no, i think your teacher said use r₁ = x, r₂ = 2 - x

find the net force for each value of x, then put it equal to zero … that should give you a quadratic equation …

what do you get?

 

[xxabr]

Oh, okay. That makes more sense.
Sorry, if I don't it this right away. This unit is kind of hard for me right now.
How do I find the net force for x? :/

 

[tiny-tim]

hi xxabr!

(just got up :zzz: …)

How do I find the net force for x? :/

"net force" is another name for "total force" …

so just add the two forces (remember, one will be minus, since they're in opposite directions) …

what do you get? ​

 

[xxabr]

Oh okay. I actually made a mistake it's not 2-x it's (2-x)². I think I got the answer. I did:

Fe₁= kq1q3/r²
= (9x10⁹)(1.6x10^-5)(3.0x10^-6)/x²
=0.432/x²

Fe₂= kq2q3/r²
= (9x10⁹)(6.4x10^-5)(3.0x10^-6)/(2-x)²
= 1.728/4-4x+x²

Then I made them equal each other and cross multiplied:
0.432/x² = 1.728/4-4x+x²

1.728x² = 1.728 - 1.728x + 0.432x²
0 = -1.296x² - 1.728x + 1.728

That's how I got the quadratic formula.
I substituted and my final answer for x was x = 0.665m.

 

[tiny-tim]

yes, that's the correct method and result, except you could have shortened it, which would have made it a lot easier …

you know that some of the factors are the same on each side of the equation, so you could have canceled them before doing the final equation (or you could even have left them out completely, other than as symbols such as q₃) …

in this case, you know that one charge is 4 times the other, so one force will be 4 times (2-x)²/x² times the other,

so (2 - x)² = 4x², or 3x² + 4x - 4 = 0, or (3x - 2)(x + 2) = 0,

so x = 2/3 (0.667 m ) …

that even gives you the other equilibrium position (which the question didn't ask for), at position -2, ie 2 m outside

btw, you can easily see that 2/3 m must be correct … that's twice as close to one as to the other, which means the effect is 4 times as strong, which balances out one charge being 4 times greater (and you could have done it that way if the teacher hadn't told you to use the more general method)

 

[xxabr]

Wow, yeah, that is easier.
Thank you soo much. (: