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!

Determining magnitude of forces on metal balls

  1. Feb 9, 2009 #1
    Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls are negatively charged. The two outside balls are separated by 20 cm and the middle ball is exactly halfway in between.The absolute value of the charge on each ball is the same, 1.79 μCoulombs

    (a) What is the magnitude of the attractive force on either outside ball due ONLY to the positively-charged middle ball?
    (b) What is the magnitude of the repulsive force on either outside ball due ONLY to the other outside ball?
    (c) What is the magnitude of the net force on either outside ball?




    Please help?
     
  2. jcsd
  3. Feb 9, 2009 #2

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    look up in your book "Coulomb's law". Don't forget to convert your cm to meters, and your microcoulombs to coulombs.
     
  4. Feb 9, 2009 #3
    F = k(q_a)(q_b)/r^2

    I know so far I should set it up like..

    F = 8.99E9

    but I don't know what the other terms apply to...help?
     
  5. Feb 9, 2009 #4

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    k=8.99e9

    For part (a):
    q_a is the charge of one of the outside balls (-1.79 micro Coulombs)
    q_b is the charge of the inside ball (1.79 micro Coulombs)
    r is the distance between them (20 cm)

    if you use 8.99e9 for k, the formula is expecting Coulombs, not microcoulombs, and meters, not centimeters. So convert these. Also, look up the units for k and include them. Then you can cancel your units. You know force has units of Newtons, which is kg * m/s^2, so your units should cancel to that.

    Parts (b) and (c) are similar.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook