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: Electric problem

  1. Jan 19, 2009 #1
    Note: ^i means carrot on top of variable

    1. The problem statement, all variables and given/known data
    A particle with charge 4 µC is located on the x-axis at the point 10 cm, and a second particle with charge -3 µC is placed on the x-axis at 4 cm. What is the magnitude of the total electrostatic force on a third particle with a charge 4 µC placed on the x-axis at -2 cm? Answer in units of N.

    3. The attempt at a solution
    1) I converted all my variables to µC = C , and cm = m
    q1 = 4 µC = 4e-6
    q2 = -3 µC = -3e-6
    q3 = 4 µC = 4e-6
    x1 = 10 cm = 0.1 m
    x2 = 4 cm = 0.04 m
    x3 = -2 cm = -0.02 m

    2 By using Coulom's law in vector form:
    F13= ke(q1q3/r2)^r13
    where ^r13 is a unit vector directed from q1 to q3; i.e., ->r13 = ->r3 - ->r1

    x13 = x3 - x1
    (-0.02m)-(0.1m) = -0.12m
    x23 = x3 - x2
    (-0.02m)-(0.04) = -0.06 m

    ^x13 = x3-x1/ √(x3-x1)^2 = -1 or -^i
    ^x23 = x3-x2/ √(x3-x2)^2 = 1 or +^i

    Since the forces are collinear, the force on the third particle is the algrebracis sum of the forces between the first and third and the second and third particles.

    ->F = ->F13 + ->F23
    = ke [q1/ r213 X ^r13 + q2 / r223 x^r13 ] q3

    After putting in the variables, I got 33.9444444445 N as my answer, but was wrong. What did i do wrong?
    Last edited: Jan 19, 2009
  2. jcsd
  3. Jan 20, 2009 #2


    User Avatar
    Homework Helper

    First of all find the force between 4micro C at -2 cm and 4 micro C at 10 cm. It is repulsive force. Next find the force between 4micro C at -2 cm and -3 micro C at 4 cm. It is attractive force. Now find the net electrostatic force on the third charge.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook