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Place a 3rd charge so net force is 0 
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#1
Mar2612, 06:18 PM

P: 5

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/r_{1to3}^{2} = Q2/r_{2to3}^{2} Q1/(x+0.03m)^{2}=Q3/x^{2} Q1*x^{2} = Q3*(x+0.03m)^{2} 2q*x^{2} / q = (x+0.03m)^{2} 2x^{2} = (x+0.03m)^{2} 2x^{2} = x^{2} + 0.06x + 0.0009 0=3x^{2} + 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? 


#2
Mar2612, 07:09 PM

HW Helper
P: 2,316

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 = x^{2}  0.06x  0.0009  slightly different to your equation. 


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