# Coulomb's Law particle charge

1. Jan 13, 2006

### Punchlinegirl

A particle with charge 5$$\mu$$C is located on the x-axis at the point -6 cm, and a second particle with charge -8 $$\mu$$C is placed on the x-axis at 10 cm. The Coulomb constant is 8.9875 x 10^9 Nm^2/C^2. What is the magnitude of the total electrostatic force on a third particle with charge 9$$\mu$$C placed on the x-axis at -2 cm. Answer in units of N.
First I converted everything.
-8$$\mu$$C= -8 x 10^-6 C
5 $$\mu$$C= 5 x 10^-6 C
9 $$\mu$$C= 9 x 10^-6 C
-6 cm= -.06 m
-2 cm= -.02 m
10 cm=.10 m
then I drew a free body diagram for the forces on the third particle and found that both forces point to the right, since the first particle has a positive charge, and the second has a negative charge.
Then I used Coulomb's Law,
For the first charge,
$$F= 1/4\pi \epsilon * (5 x 10^-6 * 9 x 10^-6) / (.04)^2$$
Simplifying this gave me $$1/4\pi \epsilon * 2.81 x 10^-8$$
Plugging in 8.9875 x 10^9 for E and solving gave me 2.48 x 10-19.
Then I did the same thing for the next charge.
$$F= 1/4\pi \epsilon * (-8 x 10^-6 * 9 x 10^-6)/ (.12)^2$$
so $$1/4\pi \epsilon * 5 x 10^-9$$
Plugging in for E and solving gave me 4.43 x 10^-20.
I then added these together, and got 2.923 x 10^-19 N, which wasn't right.

2. Jan 13, 2006

### Staff: Mentor

I haven't checked your arithmetic, but I suspect the problem is here:
Realize that the Coulomb constant $$k = 1/(4\pi \epsilon_0)$$.

Realize that that number is k, not $$\epsilon_0$$.

Other than that, your solution looks OK.

3. Jan 13, 2006

### Punchlinegirl

Thanks so much Doc Al. Now I'm finally getting some right answers on my homework.