Solving for Net Force of Charge Q on Two Charges q1 & q2

AI Thread Summary
The discussion revolves around determining the position of charge Q such that its net force is zero due to two other charges, q1 and q2, placed along a horizontal axis. The user initially formulates the equation for net force but struggles with the signs and simplification, leading to confusion about the correct expression. After some back-and-forth, it is clarified that the signs in the equations were mismanaged, and the correct form should yield a solvable quadratic equation. The user acknowledges the correction and expresses gratitude for the clarification. The focus remains on accurately solving for the distance x in relation to L.
kashmirekat
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Hello all,

I have two charges, q1 & q2, along a horizontal axis of length L. I supposed to determine the length at which another charge, Q, can be placed so that its net force is zero, other than infinitely away.

I'm using the equation:
F = kq1Q / r + kq2Q / r
-- > kq1Q / (L+x)^2 = -( kq2Q / x^2 ) where x is the distance the point is away from L.

I substitute my #s and I get the equation:
(8 / (l^2 + 2xL + x^2)) = -(2/x^2)
and then I get:
8x^2 = -2(L^2 + 2xL + x^2)
-4x^2 = L^2 + 2xL + x^2
0 = L^2 + 2xL + 5x^2
Is this right? I cannot seem to solve for x.

I thought I initially had it, but reworked through my math and realized I forgot to have 2q negative. What I initally got was:
L^2 + 2xL - 3x^2
I can solve that by 'unfoiling' easily, but, as I previously mentioned, the math isn't correct to get that equation.

Thank you for your help and have a wonderful day.
 
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What values are you using for q1 and q2?
 
I'm using 8 for q1 and -2 for q2.
 
Originally posted by kashmirekat

I thought I initially had it, but reworked through my math and realized I forgot to have 2q negative. What I initally got was:
L^2 + 2xL - 3x^2
I can solve that by 'unfoiling' easily, but, as I previously mentioned, the math isn't correct to get that equation.
What makes you think the math isn't correct? Looks right to me.
 
I start with this equation:
(8 / (L^2 + 2xL + x^2)) = -(-2/x^2), cross multiply
8x^2 = -2(L^2 + 2xL + x^2), divide by -2
-4x^2 = L^2 + 2xL + x^2, add -4x^2
0 = L^2 + 2xL + 5x^2, not L^2 + 2xL - 3x^2 as I had originally hoped for.
So, how do you solve for x (in terms of L) for this equation:
0 = L^2 + 2xL + 5x^2
 
Originally posted by kashmirekat
I start with this equation:
(8 / (L^2 + 2xL + x^2)) = -(-2/x^2), cross multiply
8x^2 = -2(L^2 + 2xL + x^2), divide by -2
-4x^2 = L^2 + 2xL + x^2, add -4x^2
0 = L^2 + 2xL + 5x^2, not L^2 + 2xL - 3x^2 as I had originally hoped for.
You are messing up with signs. In your first equation, the minus signs cancel. Your original hope was correct. :smile:
 
I was in the middle of asking a question about the signs and your response answered it perfectly.

Thank you.
 
Last edited:

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