What is the optimal position for a third charge to achieve a net force of zero?

[max1995]

Homework Statement

. A charge 2Q is located at the origin while a second charge -Q is located at x = 1m. Find the position x where a third charge should be placed so that the net force on this charge is zero?

Homework Equations

The Attempt at a Solution

I know that the charge must be placed further away from the origin than -Q and net force of 0=k*2Q*Q₃/(r₁)² + k*-Q*Q₃/(r₂)²
where
k=8.99x10⁹
r₁=distance from 2Q to Q₃
r₂= distance from -Q to Q₃

and also r₁=r₂+1m

Thanks for any help

 

[max1995]

quick question, I have tried again, does the distance between the origin and the third charge equal 3.41m

 

[RUber]

Are you to assume that the third charge is also a -Q?
In that case, letting x denote the position of the third charge, your guiding equation simplifies to:
$0 = \frac{-2Q^2}{(x^2)} + \frac{Q^2}{(1-x)^2}$

 

[max1995]

Are you to assume that the third charge is also a -Q?
In that case, letting x denote the position of the third charge, your guiding equation simplifies to:
$0 = \frac{-2Q^2}{(x^2)} + \frac{Q^2}{(1-x)^2}$

I don't really know.

by solving that equation does that mean x=~2.41m?

 

[ehild]

quick question, I have tried again, does the distance between the origin and the third charge equal 1.41m

quick question, I have tried again, does the distance between the origin and the third charge equal 3.41m

That is correct for the distance of the third charge from the origin. But show your work in detail.

 

[RUber]

That (2.41) doesn't seem to solve the equation. I get about 1 instead of zero when I plug that in. 3.41 is good.

 

[ehild]

Are you to assume that the third charge is also a -Q?
In that case, letting x denote the position of the third charge, your guiding equation simplifies to:
$0 = \frac{-2Q^2}{(x^2)} + \frac{Q^2}{(1-x)^2}$

No need to assume anything about the third charge.

 

[RUber]

No need to assume anything about the third charge.

Got it. It can factor out just like k.
Thanks.

 

[max1995]

Thank you all very much :)