## anonymous one year ago Can someone please check my work on this equation so far? Here is the question... Two point charges each carrying a charge of + 4.5 E – 6 C are located 4.5 meters away from each other. How strong is the electrostatic force between the two points and is this force a repulsive force or an attractive force (k = 9.0 E9 Nm2/C2)? So the values I have are as follows... F = ? k = 9.0 E9 Nm^2/C^2 q1 = 4.5 E - 6 C q2 = 4.5 E - 6 C r = 4.5 m Am I on the right track? Thanks!

1. ash2326

@SarahCatherinez yes you have got the values of each of the parameter correct. Seem to be on the right track. Go on and plug in the values in the formula for force.

2. anonymous

Great! I'll go ahead and plug in the values, if you wouldn't mind checking my work from there.

3. ash2326

yes, no problem

4. anonymous

F = (9.0 E9) * (4.5 E -6) * (4.5 E -6)/(4.5^2) F = (9.0 E9) * (4.5 E -6) * (4.5 E -6)/(20.25) F = (9.0 E9) * (4.5 E -6) * (2.2 E -7) F = .00891 N

5. ash2326

There seems to be a mistake in 3rd step.Could you check?

6. anonymous

Hm, that's odd. I just entered it into my calculator and got the same answer.

7. ash2326

yes, sorry. Your answer is close to the correct. I would show you another way you have this $F=\frac{9*10^9*(4.5 *10^{-6})^2}{4.5^2}$ $F=\frac{9*10^9*(4.5)^2 *(10^{-6})^2}{4.5^2}$ $F=\frac{9*10^9*\cancel{(4.5)^2} *(10^{-6})^2}{\cancel{4.5^2}}$ $F=9*10^9*10^{-12} N$ Do you follow this method?

8. anonymous

Oh yes, that makes sense.

9. ash2326

I think you can find the answer now, can't you?

10. anonymous

I believe so. Thank you so much for your help! :)

11. ash2326

No problem