# Net force on a charge

1. Nov 21, 2009

### johnnyies

1. The problem statement, all variables and given/known data
Three charges lie along the x -axis. The positive charge q1 = 10.0 microC is at x = 1.00 m, and the negative charge q2 = -2.00 microC is at the origin. Where must a positive charge q3 be placed on the x-axis so that the resultant force on it is zero?

Answer: x = - 0.809 m

2. Relevant equations
Columb's Law
F= k q1 q2
- - - - - -
r^2

k = 8.9875 x 10^9

3. The attempt at a solution

Force of 1 acting on 3 = - k q1 q3 / (1 - x)^2
Force of 2 acting on 3 = k q2 q3 / x^2

k q2 q3 / x^2 - k q1 q3 / (1 - x)^2 = 0

k's and q3's cancel out and I get

q2(1 - x)^2 = q1(x^2)
-2(1 - 2x + x^2) = 10x^2
-2 + 4x - 2x^2 = 10x^2
12x^2 - 4x +2 = 0

x = .167 m

Needing a bit of help in setting this one up perhaps. No solutions guide is available. Can someone have a more conceptual explanation on how to solve this one?

Last edited: Nov 21, 2009
2. Nov 21, 2009

### Staff: Mentor

Thanks for showing your work -- makes this much easier.

This line has a math error in it: q2(1 - x^2) = q1(x^2)

The term on the left should be quantity squared (you pulled the squared inside the parens. So re-write as:

$$q_2 (1-x)^2 = q_1 x^2$$

3. Nov 21, 2009

### johnnyies

oh, that's just a typing error, it doesn't change the answer from what I had originally.

Fixed and thanks.

4. Nov 21, 2009

### Staff: Mentor

Okay, then I think the issue that is left is that you double-did the negative sign for the negative charge:

Force of 1 acting on 3 = - k q1 q3 / (1 - x)^2
Force of 2 acting on 3 = k q2 q3 / x^2

You should let the sign on the charges themselves dictate whether the force is in the + or - x direction.

BTW, the book answer of -0.809m works in the equation you got to this point:

q2(1 - x)^2 = q1(x^2)

.

5. Nov 21, 2009

### johnnyies

double-did the negative? That doesn't make sense. I thought you had to arbitrarily add the negative sign cause that's the direction the force will be in?