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: Coulomb's Law problem

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Particle 1 of charge q1 = +0.76 µC and particle 2 of charge q2 = -3.0 µC, are held at separation L = 13 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the x and coordinates of particle 3?

    I'm really not sure how to proceed with this question. But I understand that the forces added together is zero.
  2. jcsd
  3. Mar 16, 2009 #2


    User Avatar
    Homework Helper

    Welcome to PF.

    OK. So what equation do you think you might use then?
  4. Mar 16, 2009 #3
    So I did:

    c/d^2 + c/d^2 =0

    put in the numbers

    (0.76*10^-6)/d^2 = (3*10^-6)/(d+13)^2

    simplify taking out the fractions

    (0.76*10^-6)(d+13)^2 = d^2(3*10^-6)

    expand the bracket

    (0.76*10^-6)(d^2+26d+169) = d^2(3*10^-6)

    eventual rearrangement

    2.24*10^-6 - 1.976*10^-5 - 1.2844*10^-4 = 0

    quadratic equation

    13 and -4.3 but 13 is not the right answer. It might be my method or calculation error, I'm really not sure.
  5. Mar 16, 2009 #4


    User Avatar
    Homework Helper

    I trust you didn't round away too much from the answer for your quadratic.

    If you think about it, ≈ 13 is an expected result, because if 1 charge is 4 times larger and twice as far away ... you'd expect something about equal to the charge separation wouldn't you?
  6. Mar 16, 2009 #5
    Hum, isn't q2 negative? I reckon you might've used it as a positive..

    F1 - F2 = 0 .. so, q1/d^2 = q2/(d+x)^2, where q2 still is negative, and I only see +3 in your calculations :)
  7. Mar 16, 2009 #6


    User Avatar
    Homework Helper

    Actually I think he's taken that into account.

    ∑F = 0 = q3*∑ E = 0 ⇒ E1 + E2 = 0

    Taking account of the sign of the charge then |E1| = |E2| satisfies the condition.

    (This of course is for x that does not lie between q1 and q2.)
  8. Mar 16, 2009 #7
    I don't quite see it, but I reckon you're right 'cause you most definitely know better than I do and it's late & I'm tired. I think I'll come back and try to see it when it's not 20 to midnight ;)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook