Place a 3rd charge so net force is 0

[jakerue]

Homework Statement

A charge of +2q is placed at the origin and a second charge of -q is placed a x = 3cm. Where can a third charge +Q be placed so that it experiences a zero force?

Homework Equations

F=kQ1Q2/r^2

The Attempt at a Solution

OK I thought I had this until the roots of the quadratic I am solving come out to be imaginary...

Q1 = 2q
Q2 = -q
Q3 = +Q

I am assuming that Q3 must be placed to the right of Q2.
the distance between Q1 and Q3 = x+0.03m and the distance between Q2 and Q3 = x

If F on Q3 = 0 then FQ1=FQ2 at this point

F1on3 = k*Q1*Q3/r^2 = F 2on3 = k*Q2*Q3/r^2 ;; cancel k and Q3

Q1/r_1to3² = Q2/r_2to3²

Q1/(x+0.03m)²=Q3/x²

Q1*x² = Q3*(x+0.03m)²
2q*x² / -q = (x+0.03m)²
-2x² = (x+0.03m)²
-2x² = x² + 0.06x + 0.0009

0=3x² + 0.06x + 0.0009

Can anyone tell me where I've gone wrong here? Is my assumption that Q3 is to the right of Q2 incorrect?

 

[PeterO]

Homework Statement

A charge of +2q is placed at the origin and a second charge of -q is placed a x = 3cm. Where can a third charge +Q be placed so that it experiences a zero force?

Homework Equations

F=kQ1Q2/r^2

The Attempt at a Solution

OK I thought I had this until the roots of the quadratic I am solving come out to be imaginary...

Q1 = 2q
Q2 = -q
Q3 = +Q

I am assuming that Q3 must be placed to the right of Q2.
the distance between Q1 and Q3 = x+0.03m and the distance between Q2 and Q3 = x

If F on Q3 = 0 then FQ1=FQ2 at this point

F1on3 = k*Q1*Q3/r^2 = F 2on3 = k*Q2*Q3/r^2 ;; cancel k and Q3

Q1/r_1to3² = Q2/r_2to3²

Q1/(x+0.03m)²=Q3/x²

Q1*x² = Q3*(x+0.03m)²
2q*x² / -q = (x+0.03m)²
-2x² = (x+0.03m)²
-2x² = x² + 0.06x + 0.0009

0=3x² + 0.06x + 0.0009

Can anyone tell me where I've gone wrong here? Is my assumption that Q3 is to the right of Q2 incorrect?

Your initial assumption is correct.

I would be looking as follows:

The charges of 2q and q [never mind the signs] means the force form the forst charge is potentially twice the size.
The way to balance that is to have it √2 times as far from your Q

so in your nomenclature

x + 0.03 = √2x [remember that is only √2 the x is not under the √ sign]

That leads to 0 = x² - 0.06x - 0.0009 - slightly different to your equation.