1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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
  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;
    K (1)(-0.35x10^-9)

    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.

    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)
    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)
  13. Sep 23, 2009 #12
    according to my most recent calculations, r = -1.79 x 10^-28m

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook