Given two charges where to put thrid charge for zero net force

[msbell1]

Homework Statement

Two charges, q1 and q2, are separated by a distance r. Find the position at which a third charge, q3, can be placed so that the net electrostatic force on it is zero.

Homework Equations

Coulomb's Law:
F = (kq₁q₂)/r₁₂² * (r₁₂hat)

The Attempt at a Solution

I was trying to solve this for the general case, without making any assumptions about the relative signs of q1 and q2. However, this kind of turns into a mess (for me at least), so I changed my strategy. I now consider two cases. One is where q1q2>0 (they have the same sign. In this case, I know that q3 will have to be placed somewhere between q1 and q2. This gives me (after assigning q1 to the origin, and q2 to x=r, and placing q3 at position x),

(kq1q3)/x^2 = (kq2q3)/(r-x)^2

Solving this for x, I find:
x = [(q1 (+or-) sqrt(q1q2))/(q1-q2)]*r

This looks kind of ok, except, what happens when q1=q2? That doesn't make any sense. It would look better if it was (q1+q2) in the denominator. Then I thought maybe I messed up a sign in my original equation, so I tried

(kq1q3)/x^2 = -(kq2q3)/(r-x)^2

(I just put a minus sign in front of the term on the RHS.) This equation actually seems like it should be appropriate when q1q2<0 (the have opposite signs). Solving this gives:

x = [(q1 (+or-) sqrt(-q1q2))/(q1+q2)]r,

and this equation has perfect behavior for the case where it is valid, when q1q2<0.

Can anyone see what I did wrong in the first solution? I know if q1=q2 I should get x=r/2, not x=∞. Thanks!

 

[Delphi51]

(kq1q3)/x^2 = (kq2q3)/(r-x)^2 [1]

Solving this for x, I find:
x = [(q1 (+or-) sqrt(q1q2))/(q1-q2)]*r [2]

I like this part.
In the middle step when you cross multiply and put in the form ax² + bx + c = 0
you can see that when q1 = q2, it is not a quadratic so for that case you would have to do something other than [2] to solve it. Check to see if it works out to x = r/2.
Maybe it works for q1 and q2 having opposite signs, too.

 

[msbell1]

Thank you very much! I see that now! If I try this when q2 = -q1, I find
x = [(1 (+or-) sqrt(-1))/2]*r. Since this is imaginary, I guess this indicates that there is no solution between q1 and q2 when their signs are opposite, which is true.

 

[Delphi51]

Most welcome! Did you try q1 = 4, q2 = -1, q3 = 1? That should show if it works for q3 to the right of q2.