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Where should the proton be placed to balance F?

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A point charge q = -0.5nC is fixed at the origin. Where must a proton be placed in order for the electrical force acting on it to be exactly opposite its weight?


    2. Relevant equations

    F=
    k(q1)(q2)
    ------------
    r^2


    3. The attempt at a solution
    Trying to solve for distance (r), but coming across a few problems;

    If the force acting on it is exactly opposite its weight, does this make F=0?

    should I be substituting 0 for F?
    Where would I substitute -0.5? in q1 or q2?
    And what would I place in the other?

    ???
     
  2. jcsd
  3. Sep 21, 2009 #2
    A tips for the force: When you break out r you will need to take the square root, the value inside must be positive to yield a non-imaginary number. What does this say about the force?
    It can't be zero, because then you would have to divide by zero, which gives a non-defined answer.

    For the charges, it doesn't matter if the proton is q1 or q2, the same goes for the point charge. The expression won't change.
     
  4. Sep 21, 2009 #3
    KQ1Q2/r^2=mg
     
  5. Sep 21, 2009 #4
    Anden, I substitute 1 for q1 (the proton) and -0.35 x 10^-9 for q2 (the point charge)?

    so I now have;
    F=
    K (1)(-0.35x10^-9)
    ---------------
    r^2

    Dadface, were you saying I should use F=m.g to find the value of force?
    g would be 9.8,
    what would m be?

    would I use the mass of an electron (9.1×10^-31kg)
     
  6. Sep 21, 2009 #5
    the equation I now have is;
    F= (9.1x10^-31)x(9.8). This equals 8.918x10^-30.
    therefore;
    8.918x10^-30=
    (9x10^9)x(1)(-0.35x10^-9)
    ---------------
    r^2

    Then solve for r?
     
  7. Sep 21, 2009 #6
    A proton does not have the charge 1 C, that is the charge of 6,215 * 10^18 protons.

    Dadface equation is almost right, with one modification "mg" needs to be replaced by "-mg", this is needed to ensure a positive result inside the square root.

    The mass you must is the one for the proton, its mass is roughly 1840 times bigger than the electrons
     
  8. Sep 21, 2009 #7
    so F=(9.81)x(1.672x10^-27) => F=1.64x10^-26

    the last thing I'm confused about is what to substitute for q1 and q1?
    I know they are the point charges, and that one of them = -0.35x10^-9.
    How do I figure out what the other point charge is?
    Once I have that I should be good to go.

    =]
     
  9. Sep 21, 2009 #8
    The charge on the proton is 1.6 times ten to the power of minus nineteen Coulombs
     
    Last edited: Sep 21, 2009
  10. Sep 21, 2009 #9
    Protons and electrons have the same charge, but with different signs. Protons are positive 1 e, and electrons are negative 1 e (-1 e). I gave you a hint for the charge of the proton, if 1 C equals 6,242E18 Protons (The value I posted above is wrong, sorry about that), what is the charge of one proton?
     
  11. Sep 21, 2009 #10
    The equation would now read;
    (1.64x10^-26) = (8.99x10^9) (1.6x10^-19)(-0.35x10^-9)
    ------------​
    r^2​
    is that right?

    rearrange the equation to make r the subject.

    solve the equation, giving me;
    r= -4.148x10^-46m
    ???

    Hopefully I have it right. AT least I now have some sort of clue on how to solve it at the worst.
    Thanks so much for the incredible patience Anden and Dadface.
    I only see my physics teacher every thursday so if I get stuck on something it drives me insane.
     
  12. Sep 23, 2009 #11
    still a bit unsure on this;
    q1 (the electron) has a charge of -3.5 x 10^-9 C
    q2 (the proton) has a charge of 1.6 x 10^-19 C

    to find F;
    F = m.g
    m of an electron is; 9.1 x 10^-31
    g is; 9.8

    therefore F= 8.9 x 10^-30 N

    the equation will now be;
    8.9 x 10^-30=
    9 x 10^9 x (-3.5 x 10^-9)(1.6 x 10^-19)
    ----------------------​
    r^2​
     
  13. Sep 23, 2009 #12
    according to my most recent calculations, r = -1.79 x 10^-28m

    ?????
     
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