Coulomb's Law

[kashmirekat]

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.

[Doc Al]

What values are you using for q1 and q2?

[kashmirekat]

I'm using 8 for q1 and -2 for q2.

[Doc Al]

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.

[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.
So, how do you solve for x (in terms of L) for this equation:
0 = L^2 + 2xL + 5x^2

[Doc Al]

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. [:)]

[kashmirekat]

I was in the middle of asking a question about the signs and your response answered it perfectly.

Thank you.