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Quick question on sign convention for Work in Columbs

  1. Dec 9, 2006 #1
    1. The problem statement, all variables and given/known data

    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa06 [Broken]

    2. Relevant equations

    W =F*D

    3. The attempt at a solution

    I have it solved, now i thought - would be if i put work into the thing to move it but looking at it since both of them are postive charge they are gona wana repel each other to begin with so why is it not +
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Dec 9, 2006 #2
    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06 [Broken]

    Why am i stuck on 14 and 15

    The equation that i would use would be F = Kq1Q2 / R^2 correct? and all it is is just plug and chug but i am not getting the answer
     
    Last edited by a moderator: May 2, 2017
  4. Dec 9, 2006 #3

    OlderDan

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    #14 has a variable force acting the whole time Q1 is approaching Q2. This is not a simple matter of finding one force. Fortunately, the problem has been stated in terms of energy, which is a big hint about how it should be approached. What happens to the potential energy of Q1 as it approaches Q2?

    For #15 you need to think about the relationship between voltage across a capacitor and the electrical quantity that causes that voltage. What is the same for the three capacitors?
     
    Last edited by a moderator: May 2, 2017
  5. Dec 9, 2006 #4
    All right, can u explain this
    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa06 [Broken]

    Question 15


    I assumed it would be positve, so if i do right hand rule, i have force is up so Electric field would have to point down CorrecT? so why is it negative
     
    Last edited by a moderator: May 2, 2017
  6. Dec 10, 2006 #5

    OlderDan

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    There is no electric field in the problem. The charge q does produce an electric field, but that field has nothng to do with the forces acting on q. You used the right hand rule correctly and found that the force on a positve charge would be upward at the position shown in the diagram. Clearly the force at that point must be downward, so the charge cannot be postive and must be negative.
     
    Last edited by a moderator: May 2, 2017
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