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Point charge Exam review question

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Three charges are placed along a meter stick. One charge, q1, is placed at the 12.1cm mark. A second charge, q2, is placed at the 27.2cm mark. The third charge, q3, is placed at the 48.5cm mark. Charge q1 is equal to +171μC. What must be the sign and magnitude of charge q3 such that the total electrical forces on charge q2 add to zero?

    2. Relevant equations

    Coulomb's law

    3. The attempt at a solution

    I tried equating the Electrostatic force of q1 on q2 to the electrostatic force of q3 on q2, but I don't believe it came out right.

    8.99 X 10^9 Nm^2/C^2 * (171μC*q2/15.1^2)=-8.99 X 10^9 Nm^2/C^2 * (q2*q3/21.3^2)
     
  2. jcsd
  3. Feb 7, 2012 #2

    Doc Al

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    Staff: Mentor

    Looks like you have the right idea. What must the sign of q3 be?

    Get rid of the minus sign in your equation, then solve for q3. Cancel everything that you can first.
     
  4. Feb 7, 2012 #3
    I stared at a while and realized I can cancel q2 and 8.99 X 10^9 Nm^2/C^2. I came up with q3= -340.25μC. I realized that the only way that the charges could balance out is if q3 is positive making the answer 340.25μC. Does this sound right?
     
  5. Feb 7, 2012 #4

    Doc Al

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    Staff: Mentor

    Good. But get rid of the minus sign in your initial equation.

    (The first thing you should do is determine the sign of the third charge. Then use your equation to find the magnitude of that charge.)
     
  6. Feb 7, 2012 #5
    Yeah, the minus sign came from me setting it up like this:

    8.99 X 10^9 Nm^2/C^2 * (171μC*q2/15.1^2)+8.99 X 10^9 Nm^2/C^2 * (q2*q3/21.3^2)=0.

    I guess that q2 would be a negative charge, which I think would make the math correct. I'll have to keep an eye on that to make sure my answer is correct to the situation with regard to the sign!

    Thanks for the help!!
     
  7. Feb 7, 2012 #6

    Doc Al

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    Staff: Mentor

    Yes, I know what you were trying to do. Setting F1 + F2 = 0 is the right idea, but you have to be careful when you write those forces in terms of the charges.

    If you treat q3 as a positive charge, then the force it exerts on q2 (assumed positive) will be given by F2 = -kq2q3/d2, where the minus sign indicates that the force points to the left--opposite to the force from q1. Then when you set up your equation and solve for q3 you'll get a positive answer.

    Personally, I would just set the magnitudes of the forces equal, and use that to solve for the magnitude of the charge. (You already know what the sign of the charges must be.)
    No, it has nothing to do with q2. Remember, q2 dropped out.
     
  8. Feb 7, 2012 #7
    Oh, ok. Solving for the magnitude does simplify things.
     
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