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Homework Help: Coulombs law problem

  1. Jan 20, 2005 #1
    Two charges, q1 and q2, are held a fixed distance d apart. a. Find the strength of the electric force that acts on q1. Assume that q1=q2=21.3 micro COulombs and d=1.52 m. b. A third charge q3=21.3 micro COulombs is brought in and placed as shown. Find the stregnth of the electric force on q1 now.
    My best attempt at showing the diagrams:
    a.
    q1----------q2


    b.
    q1----------q2
    __-----------
    ___--------
    ____-----
    _____q3
    b is an equilateral triangle if you cant figure out by the diagram with all sides a distance d. edit: i see the my attemtp at diagrams isnt working too well buts the general idea

    I got part a with Fe=1.77 N
    I don't know what to do for part b other than its some application of Fe=kq1q2/r^2 . Thanks for the help
     
    Last edited: Jan 20, 2005
  2. jcsd
  3. Jan 20, 2005 #2
    Part a looks alright
    For part B
    Hint:
    The net force acting on q1 is the vector sum of the individual forces caused by the other two charges.
     
  4. Jan 20, 2005 #3
    thanks i think i got it
    3.07 N sound about right
     
  5. Jan 20, 2005 #4
    yep i got same ting
     
  6. Jan 20, 2005 #5

    DB

    User Avatar

    I have a simple question: how do you convert Micro Coulombs to coulombs? (I couldn't find it on the net)
    Once I do that I use

    [tex]\frac{\mid q_1q_2\mid}{4(1.52^2)*8.854*10^{-12}*\pi}[/tex]

    right?

    P.S you did say that q1*q2 was 21.3 mC right?
     
  7. Jan 20, 2005 #6
    micro => 10^-6 (has the greek symbo "mu" )
    => 1 micro Coulomb = 10^-6 Coulombs
     
    Last edited: Jan 20, 2005
  8. Jan 20, 2005 #7

    DB

    User Avatar

    Thanks, so 21.3 mC = 2.13e-5 C
    But what have I done wrong here?

    [tex]F=\frac{\mid 2.13*10^{-5}\mid}{4(1.52^2)*8.854*10^{-12}*\pi}[/tex]

    [tex]F=\frac{\mid 2.13*10^{-5}\mid}{2.57061216 * 10^{-10}}[/tex]

    It doesnt come to 1.77 N lol
     
  9. Jan 20, 2005 #8
    look at the equation of coulombs law that you posted... it says q1*q2 but in this case q1=q2... so it should be (2.13E-5)^2
    you forgot to include the second charge :wink:
     
  10. Jan 20, 2005 #9

    DB

    User Avatar

    Awsome, my carelessness once again...

    [tex]F=\frac{\mid 4.5369*10^{-10}\mid}{2.57061216 * 10^{-10}}\approx1.77 N[/tex]

    Thanks
     
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