# 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