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: 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


    3. The attempt at a solution

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


    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..


    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

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