Mass of a particle help please

1. Sep 16, 2006

rob24

hey guys, i have been trying to work this problem out but for some reason, my answer is different from the books answer on the back.
the problem is as follow:
Two points charges of 30 nC and -40nC are held fixed on an X axis, at the origin and at X=72 cm, respectively. A particle with a charge 42 micro C is released from rest at X=28 cm. if the initial acceleration of the particle has a magnitude of 100 Km/s^2, what is the particle's mass?????

the way i was solving it:
let q1=30nC and q2= -40nC q3=42 Mic C X=72 cm
F13 = (8.99*10^9*(30*10^-9)*(42*10^-6))/(28*10^-2)^2= 0.144
F23 = (8.99*10^9*(40*10^-9)*(42*10^-6))/(44*10^-2)^2= 0.078
F12 = (8.99*10^9*(30*10^-9)*(40*10^-9))/(72*10^-2)^2=2.08*10^-5

so Fnet= -F13+F23+F12 and then F=ma so m=F/a.

Last edited: Sep 16, 2006
2. Sep 16, 2006

neutrino

Fnet on what? What's the net force acting on the third charge (42 microC)?

3. Sep 16, 2006

rob24

I did that too. it's F13 and F23. so Fnet without F12 but it's still wrong.

4. Sep 16, 2006

neutrino

Are you sure you've taken into consideration the nature of the charges?

5. Sep 16, 2006

rob24

yes. please tell me if you have different idea.

6. Sep 16, 2006

neutrino

I don't think so. You haven't included the minus sign (-) of the negative charge whlie calculating F23.

7. Sep 16, 2006

rob24

you are correct but the answer is still incorrect. i had added the - sign later after calculation. so i had - and -

Last edited: Sep 16, 2006
8. Sep 16, 2006

neutrino

What's you answer? And, btw, did you also convert Kms-2 to ms-2?

9. Sep 16, 2006

rob24

I kept getting 2.2*10^-6 but in the book he has it as 2.2*10^-5.
of course i had 100km/s^2 as 100000m/s^2

10. Sep 16, 2006

neutrino

EDITED
Oops, sorry...too late into the night for calculations. I'm getting the same answer as you. May be it's a typo in the book.

Last edited: Sep 16, 2006