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Coulomb's law and related problem

  1. Jul 12, 2011 #1
    1. The problem statement, all variables and given/known data

    The sum of two point charges is 6µC. They attract each other with a force of 0.9 N, when kept 40 cm apart in vacuum. Calculate the charges.

    2. Relevant equations



    3. The attempt at a solution
    q1+(-q2) = 6x10-6
    q1-q2=6x10-6.......I
    q1=6x10-6+q2
    According to Coulomb's law
    F=q1q2/4πε0r2
    .9=(6x10-6+q2)q2/4πε0(.4)2
    .9=(6x10-6q2+q22)x9x109/.16
    .144/9x109=6x10-6q2+q22
    144x10-3x10-9/9=6x10-6q2+q22
    q22+6x10-6q2-16x10-12=0
    (q2+8x10-6)(q2-6x10-6)=0
    q2=-8µC or q2=6µC
    The answers given in the book are q1 = 8micro Coulomb and q2= -2 microcoulomb
    I don't understand how to get this answer because when q2 = 8 microcoulomb and substituting in equation I , i get q1+8micro coulomb =6 micro coulomb so q1 should be -2 micro coulomb
    Did i miss anything in the steps? Members, please help me .
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 12, 2011 #2

    gneill

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    Staff: Mentor

    Since you're assuming that q1 is positive and that q2 is negative, you don't need to wedge in that extra minus sign. Just write q1 + q2 = 6x10-6C.
    Since one of the charges is negative the product q1*q2 will be negative, and thus the force should be negative also.

    Let f = 0.9N, r = 0.40m, Q = 6μC, [itex]k = \frac{1}{4 \pi \epsilon_o}[/itex] Then

    [itex] Q = q_1 + q_2[/itex] so that [itex] q_2 = Q - q_1[/itex]

    [itex] f = -k \frac{q_1 q_2}{r^2} = -k \frac{q_1 (Q - q_1)}{r^2}[/itex]
     
  4. Jul 12, 2011 #3
    it has been mentioned that two charges attract each other thats why i took q2 as negative. is my assumption wrong, sir?
     
  5. Jul 12, 2011 #4

    gneill

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    Staff: Mentor

    No, the assumption is fine. Oppositely signed charges attract. Either q1 or q2 must be negative (but not both!).
     
  6. Jul 13, 2011 #5
    Thanks a lot for the reply, sir. Then why i didn't get the correct answer when i framed the equation q1-q2= 6 micro coulomb, taking q1 as positive and q2 is negative. I got the correct answer when i proceeded as suggested by you. But whats wrong with my assumption?
     
  7. Jul 13, 2011 #6

    gneill

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    Staff: Mentor

    q1 - q2 is the difference between the charges, not the sum of the charges.
     
  8. Jul 13, 2011 #7
    I got it sir. Thanks a lot for patiently helping me.
     
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