# Electric Forces

1. Oct 5, 2009

### luap12

1. The problem statement, all variables and given/known data[http://img44.imageshack.us/img44/1104/58339946.jpg [Broken]][/URL]

2. Relevant equations [F=k*abs(q1)*abs(q2)/r^2]

3. I understand the equation, but I am having trouble with my signs and also how to include the distance in the equation. I know that the force can't be zero between the charges, and I think it has to be farthest away from the larger positive charge to make them equal to zero. The forces point in the same direction because they are opposite.]
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 4, 2017
2. Oct 5, 2009

### rl.bhat

Consider a point at a distance x from -q toward right, and find the forces on a unit positive charge due to two charges. Equate them to get the value of x.

3. Oct 5, 2009

### luap12

I don't understand what you mean by that... I think I'm over-thinking this.

4. Oct 5, 2009

### rl.bhat

Force on unit positive charge due to -q at a distance x*1o^-2 m from it is....?
Force on unit positive charge due to 2q at a distance (10 + x)*10^-2 m is....?
Since net force is zero, equate them to find x.

5. Oct 5, 2009

### luap12

Thanks very much, I got it now.

((k*q)/x^2)-((k*2q)/(x+10)^2)=0, solving for x gives 24.142cm. Accounting for the distance of 5cm on the x axis give the point to be 29.142cm. This was the correct answer. Thanks!