anonymous 4 years ago A proton orbits a 1.0-cm-diameter metal ball 1.30 above the surface. The orbital period is 1.60 . what is the charge on the ball?

1. anonymous

i've done the calculations over and over and over and even did it without any rounding off. it's an online assignment, i'm about to tell my instructor that the answer it has is wrong cuz i don't see anything wrong in my calculations. all the units cancel and i answer with the unit it's asking for. anyone wanna verify? $F=kq _{proton}q _{ball}/r ^{2}$ $F=ma=mr \omega ^{2}=mr(2\pi/T)^{2}$ m=1.6*10^-16kg solve for q_ball. what am i doing wrong >:O

2. anonymous

copy and paste took out units... that's 1.3mm and T is 1.6microseconds. i keep getting 4.28*10^-12 C

3. anonymous

$m=1.6*10^{-27}kg$ (for Proton)

4. anonymous

thx, that's what i did, i typed it wrong up there...so i still don't know what the problem is

5. anonymous

are you there elica85?

6. anonymous

yes!

7. anonymous

so let us solve this

8. anonymous

can post the procedure through which you found F?

9. anonymous

it was on cramster http://www.cramster.com/solution/solution/1212303

10. anonymous

it's the exact same problem since it's the solution for my text just different numbers. but i followed each step

11. anonymous

ok ok. then there is one thing which could be the cause of wrong solutions. may be you typed wrong values in doing calculations. isn't it?

12. anonymous

i checked so many times. T=1-6*10^-6s r=1.3*10^-3+.005=.0063m m=1.6*10^-27kg after setting each F equal to each other to solve for q_ball... $q _{ball}=(4\pi mr )/T ^{2}*(4\pi \epsilon _{0}r ^{2})/q _{proton}$ =(4pi(1.6*10^-27)*.0063)/(1.6*10^-6)^2 * (4pi*8.85*10^-12*.0063^2)/(1.6*10^-19)

13. anonymous

$q=2.029*10^{-11}$this is my answer.

14. anonymous

you are making a complex. just relax and be easy. first find F and then plug it in law and then find q. This complex suits to professional. not for beginners like you and me.

15. anonymous

all my chances were used up so it displayed the correct answer as being -4.47*10^-12 so i was close. i even tried with neg sign too :(

16. anonymous

why -ve sign. huh? it is proton not electron. you first find force and reply me with your answer.

17. anonymous

negative cuz the ball must be neg if it attracts the proton to orbit around it

18. anonymous

no no. it is the charge of the ball that you want to find. while charge on proton is +ve. there are only two qs. one for ball and other for proton. for ball it is required, so no sign meant for it. for proton it is always +ve. Did you get my knowledge?

19. anonymous

i think the question was probably to test the obvious. o well, i tried. if i get a B just for this one problem, i'll talk to the instructor cuz what i did should have worked. thx!

20. anonymous

ok. as you wish and good luck. but kindly make your solution as easy as possible.