Find Net Force Neutral Point b/w Q1 & Q2 Charges

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SUMMARY

The discussion focuses on determining the position of a third charge, Q3, that experiences no net force between two charges, Q1 = 6.00x10^-6 C and Q2 = -7.00x10^-6 C, separated by 8.00 cm. The user derived the equation Fe=kq1q2/r^2 and solved for x, resulting in roots of 0.998459 m and -0.0384593 m. The analysis confirmed that Q3 must be positioned to the left of Q1 to achieve equilibrium, leading to the conclusion that the correct placement is at -0.998459 m, which should be reported as a positive distance if Q3 is to the right of Q1.

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Sean1218
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Homework Statement



A charge, Q2 = -7.00x10^-6 C, is 8.00 cm to the right of charge Q1 = 6.00x10^-6 C. Where can a third charge be placed, along the line connecting Q1 and Q2, such that it experiences no net force? Give distances relative to Q1 and use a plus sign if the third charge is to the right of Q1

Homework Equations



Fe=kq1q2/r^2

The Attempt at a Solution



I did q1/x^2 = q2/(x+0.08)^2 and solved for x. I got a quadratic which gave me roots 0.998459 and -0.0384593. This is assuming that the order is Q3 then Q1 then Q2.

If I do it with the order as Q1 - Q2 - Q3, then I get two negative roots (the more negative one still matches up with 0.998459 if I take into account the 0.08 between Q1 and Q2).

So, would my answer be -0.998459 m because Q3 is to the left of Q1 and the question said to make it positive is Q3 is to the right? I have one try left to get this right, but I'm not confident this is correct.
 
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Looks right to me.
 

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