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Coulomb's law and two point charges question

  1. Feb 22, 2015 #1
    1. two point charges q1 and q2 with a combined charge of 20 micro coulombs are placed 3m apart. If one repels the other with a force of 0.075N calculate;

    the value of q1 and q2



    2. k=8.987x10^9
    F=kxq1xq2/r^2
    E=kQ/r^2
    E=f/q



    3. I've tried to work out the charge by working out the electric field strength with respect to one charge, then using the E=f/q and rearranging for q, this gave a value of

    q1=3.755x10^-6

    I then tried to sub this into coulombs force and rearrange for q2 but the value was faar too high.

    I then tried to the same method and subtracting q1 the combined charge, but I'm certain this is not the method

    Any help would be greatly appreciated
     
  2. jcsd
  3. Feb 22, 2015 #2

    Bystander

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    You've been give a separation distance. You've been given a resultant force between the charges. You've been given a value for one of the charges. It's not necessary to use every equation that's been given you.
     
  4. Feb 22, 2015 #3

    gneill

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    Hi deeko1987, Welcome to Physics Forums.

    Can you write two separate equations that describe the given conditions?
     
  5. Feb 22, 2015 #4
    Hi thanks for the welcome,

    I have used coulomb's force to work out the value of q1 x q2 which = 7.51x10^-11

    and I know the combined value of q1+q2 = 20x10^-6

    I'm not sure where I would go from there as the post above says its not necessary to use all the equations. Would it be wise to possibly try to get a simultaneous equations?
     
  6. Feb 22, 2015 #5

    gneill

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    Be careful with the sign of the force! Are the charges attracting or repelling each other? What does that tell you about the product of the charges?

    Note that you have already just written two simultaneous equations!
     
  7. Feb 22, 2015 #6
    the charges must be positive because the combined charges (q1 + q2) is positive. and they are repelling which off the top of my head is a positive value and a negative suggests they are attracted (will double check).

    Still struggling with this equation but I'll work through it. Its most likely one of those things where It'll be obvious once I get it.

    Many thanks
     
  8. Feb 22, 2015 #7
    I'm still hitting a brick wall with this one. as I have a equivalent; x+y=a value and X*Y=a value I cannot substitute or eliminate to isolate one of the variables.

    Do I need a third variable or formula?

    any help would be appreciated

    Deeko
     
  9. Feb 22, 2015 #8

    gneill

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    Why not? x + y = z, so x = z - y. Right?
     
  10. Feb 23, 2015 #9
    From the question, ## q_1+q_2=20μC => q_2=20*10^{-6} - q_1 ##
    From the given data, obtain ## q_1*q_2 ## ......
    Now substitute for ##q_2## in the second equation from the first.
    Solve the quadratic equation to obtain ##q_1## and hence ##q_2##
     
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