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Determining particle charge

  • Thread starter scholio
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  • #1
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Homework Statement



a charged object having 25 microcoulombs on it breaks into two pieces. if the force of repulsion between the two pieces i 1000newtons when the pieces are 3cm apart, what is the charge on each piece? assume pieces act as a point charges

Homework Equations



E = F/qtest = k(qa)/r^2 where F is the attractive/repulsive force

F = k(qa)(qb)/r^2 where k is 9*10^9, qa is charge of one piece, qb is charge of other piece, r is the radius between pieces


The Attempt at a Solution



firstly, do i assume the parts broken into are the same size/charge as eachother? i don't think so, correct?

basically i used the second eq and set F = 1000, r = 0.03m, k = 9*10^9 and solved for qa in terms of qb

i then used the first eq and subbed in F =1000, qtest = 25*10^-6 coulombs and set that equal to kqa/r^2, where i subbed in what i got from the second eq in for qa and solved for qb

i got 2.5*ac^-5 coulombs for qb, i used qb in the second eq and got a qa of 4*10^-6

i know i did something wrong because when i sum the charges of each piece, qa and qb, i get less than the entire charge value given (25 microcoulombs). i get a sum of 2.9*10^-5 coulombs

is my approach even close, help appreciated...
 

Answers and Replies

  • #2
G01
Homework Helper
Gold Member
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OK you starting from the right place, but I think you making this more complicated than it has to be:

[tex]F=1000N=\frac{kq_aq_b}{r^2}[/tex]

The problem is that you have two variables, q_a and q_b.

HINT: Can you plug in and expression for q_b in terms of q_a in the above equation using what you know? Think conservation of charge. If so, you should then be able to solve for q_a, and thus, q_b using the expression you used previously.
 
  • #3
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how about substituting in (25*10^-6 - q_a) in for q_b where 25*10^-6 is 25 microcoulombs?

when i subbed that in for q_b the eq you gave produced a quadratic equation in terms of q_a, i did the quadratic formula and got 225415.9 and -415.9

i'm not too sure on units because it is coulombs that is impossible... help?

__________%%%%%%%%%%%%%%%%%%%%%%%________________

actually i incorrectly calculated the quadratic eq, for q_a i got 1.99*10^-5 and 5.05*10^-6 coulombs, which one do i use for q_a? i'm guessing the second one...

well i chose correctly because when i subbed that value back in and solved for b i got 1.98*10^5
so q_a + q_b = 2.5*10^-5 =25*10^-6 = 25 microcoulombs
1.98*10^-5 + 5.05*10^-6 = 2.5*10^-5

YES correct!!

Thanks so much G01
 
Last edited:
  • #4
G01
Homework Helper
Gold Member
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Anytime. Good Job! :smile:
 

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