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Place a 3rd charge so net force is 0

  • Thread starter jakerue
  • Start date
1. The problem statement, all variables and given/known data

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?


2. Relevant equations

F=kQ1Q2/r^2

3. 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/r1to32 = Q2/r2to32

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 Helper
2,425
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1. The problem statement, all variables and given/known data

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?


2. Relevant equations

F=kQ1Q2/r^2

3. 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/r1to32 = Q2/r2to32

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.
 

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