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Three Point Charges in a Row, Find Magnitude 1.5cm left of middle charge

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider three charges arranged as shown.

    + + -
    1 2 3
    What is the magnitude of the electric field
    strength at a point 1.5 cm to the left of the
    middle charge? The value of the Coulomb
    constant is 8.98755 × 10^9N  m2/C2.
    Answer in units of N/C

    q1=8.2 µC= 8.2*10^-6 C
    q2=4.5 µC = 4.5*10^-6 C
    q3=−4 µC= -4*10^-6 C

    r1,2=4.4 cm= .044m
    r2,3= 4.6 cm=.046m

    kC= 8.98755*10^9

    2. Relevant equations

    F=kC*q1q2/r^2

    3. The attempt at a solution

    so I am guessing I need to start off with something like this...

    F2,1=kC*q2q1/r2,3^s

    F2,3= kC * q2q1/r2,3^2

    and then find Fnet= F2,1 +F2,3

    One of the main things I am struggling with is the 1.5cm (.015m) left of the center point.
     
  2. jcsd
  3. Sep 6, 2011 #2
    You said in the original question that you wanted to find the electric field, but then started trying to calculate the force! I think the reason you're struggling is because you can't calculate the force of charges on a bit of chargeless space. Your logic is fairly sound otherwise!
     
  4. Sep 6, 2011 #3
    Oh lol...

    Then I use E=F/q=kq/r^2

    then do I calculate the E for each one?
    like so..


    E1=k*q1/r^2
    E2=k*q2/r^2
    E3=k*q3/r^2

    then what would r be?
     
  5. Sep 6, 2011 #4
    What do you think r would be?
     
  6. Sep 6, 2011 #5
    I am guessing the r should be different for each one...

    E1=k*q1/r^2 r=.044m -.015m=.029m
    E2=k*q2/r^2 r=.015m
    E3=k*q3/r^2 r=.046m +.015m=.061m
     
  7. Sep 6, 2011 #6
    Yes, exactly that!
     
  8. Sep 6, 2011 #7
    So now
    E1=8.76*10^7
    E2=1.8*10^8
    E3=9.66*10^6

    so, Enet=E1-E2+E3

    Did I get the signs right?
     
  9. Sep 6, 2011 #8
    Yes, good spot!
     
  10. Sep 6, 2011 #9
    Thanks so much! I got it! :)
     
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