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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0copy and paste took out units... that's 1.3mm and T is 1.6microseconds. i keep getting 4.28*10^12 C

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[m=1.6*10^{27}kg\] (for Proton)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thx, that's what i did, i typed it wrong up there...so i still don't know what the problem is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are you there elica85?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can post the procedure through which you found F?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it was on cramster http://www.cramster.com/solution/solution/1212303

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's the exact same problem since it's the solution for my text just different numbers. but i followed each step

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i checked so many times. T=16*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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[q=2.029*10^{11}\]this is my answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0all 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 :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why ve sign. huh? it is proton not electron. you first find force and reply me with your answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0negative cuz the ball must be neg if it attracts the proton to orbit around it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok. as you wish and good luck. but kindly make your solution as easy as possible.
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