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Solving Repulsive Force problem

  1. Jul 16, 2007 #1
    Hello all. I am in my second week of physics and have come across a problem I can't figure out for the life of me. I'm pretty sure i'm using the wrong formula to figure it out. Heres the question:

    Two charges, Q1 and Q2, are separated by 4·cm. The repulsive force between them is 64·N. In each case below, find the force between them if the original situation is changed as described.

    a. The magnitude of charge Q1 is reduced by 1/2.
    b. The distance between the charges is doubled?
    c. The magnitude of charge Q2 is increased by 16 and the distance between the charges is increased to 8·cm.

    Now I have been assuming that each charge is 32 N. I am using the formula kQ1Q2/r2. When I do (9x10^9)(16)(32)/16 I come up with 2.88e11 which can't possibly be the correct answer.

    Can someone point me in the correct direction to solve this problem? Thank you.
     
  2. jcsd
  3. Jul 16, 2007 #2

    danago

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    Is that supposed to read 32 C? And where did you get 32 from?

    Also, you need to covert the 4cm into meters.
     
  4. Jul 16, 2007 #3

    Dick

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    Charge isn't measured in newtons. And there is no need to find sample values to plug in. Just think. E.g. for a) how does kQ1Q2/r^2 compare with k(Q1/2)Q2/r^2.
     
  5. Jul 16, 2007 #4
    I guess I was assuming each charge was contributing 32 N of force. For a. I thought by halfing the 32 to get 16 would then make the total force 48 but that didn't work out. My teacher really didn't get into this type of problem so i'm stuck trying to figure it out myself. I've looked quite a few places on the net and can't seem to find a formula that fits this problem, even my book doesn't get into this.

    Thanks again.
     
  6. Jul 16, 2007 #5

    Dick

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    Whatever the numbers are you can still compare kQ1Q2/r^2 with k(Q1/2)Q2/r^2. They are the same in each expression. How do the products compare?
     
  7. Jul 16, 2007 #6
    I believe one is half of the other, but i'm not sure what to do with the 64 N. Do I split it between the 2 charges and plug that into the formula with .04 meters or do I need to convert the Newtons into Coulombs somehow?
     
  8. Jul 16, 2007 #7

    danago

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    You know that under the normal conditions, the force is 64N.

    So [tex]
    \frac{{kq_1 q_2 }}{{r^2 }} = 64
    [/tex]

    And under the new conditions, a force of 'F' acts.

    [tex]
    \frac{{0.5kq_1 q_2 }}{{r^2 }} = F
    [/tex]

    You can set up equations like this:

    [tex]
    \frac{{{\textstyle{{0.5kq_1 q_2 } \over {r^2 }}}}}{{{\textstyle{{kq_1 q_2 } \over {r^2 }}}}} = \frac{F}{{64}}
    [/tex]

    And solve for F, by first simplifying the fraction.
     
    Last edited: Jul 16, 2007
  9. Jul 16, 2007 #8

    Dick

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    Yes, one is one half of the other. If the original force is 64N, then what is one half of that? There is no need to convert anything into anything.
     
  10. Jul 16, 2007 #9
    What should I be using for q1 and q2? Is it not 32 for each or am I way off?
     
  11. Jul 16, 2007 #10
    Look at this all over again:

    [tex]F_e = k\frac{q_1 q_2}{r^2}[/tex]

    Where k is about:

    [tex]8.998x10^9 \frac{N m^2}{C^2}[/tex]


    Just think rationally, and plug in the distortions to the equation. Record your results. =P
     
    Last edited: Jul 16, 2007
  12. Jul 16, 2007 #11

    Dick

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    I don't know why you are fighting this. You know F=64N. That is the force on each charge. Not the force on one. You do not know q1. You do not know q2. You do not know r. And you never will. You only know the combination k*q1*q2/r^2. This makes your life easier. You've already said the correct answer. And it has a 'one half' in it.
     
  13. Jul 16, 2007 #12
    Ok so (a) is 32. Somehow I figured out that (c) is 16 and that was correct. I've tried a couple more times and I believe (b) would be 8? I'm still fuzzy on how exactly to get to that but i've never been good with this type of stuff.
     
  14. Jul 16, 2007 #13

    Dick

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    (b) is not 8. You are guessing. What is the ratio between q1*q2/r^2 and q1*q2/(r/2)^2????????????
     
  15. Jul 16, 2007 #14
    It is 1 to 2.
     
  16. Jul 16, 2007 #15

    Dick

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    Maybe the source of confusion is phrases like 'increased by 16'. In the context of this problem, this cannot mean q2->q2+16. It has to mean q2->q2*16. '16' doesn't have units on it. It has to be a multiplicative factor. AND I don't believe (c) is 16N either.
     
  17. Jul 16, 2007 #16

    Dick

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    Think again. What is 1*1/(1^2) vs 1*1/((1/2)^2). Hint: (1/2)^2=1/4 and 1^2=1.
     
    Last edited: Jul 16, 2007
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