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Homework Help: Electric field?

  1. Sep 2, 2009 #1
    Electric field???

    1. The problem statement, all variables and given/known data

    A point charge of -4 uC is located at x = 2 m, y = -2 m (called A) . A second point charge of 12 uC is located at x = 1 m, y = 2 m. (called B)

    (a) Find the magnitude and direction of the electric field at x = -1 m, y = 0. (called C)

    (b) find the angle. I know i can find angle using arctan after i find the x and y components above. so this im not too worried about.

    2. Relevant equations

    Enet = Enetx + Enety

    kq / r^3 * R where R is the vector (equation given by TA)

    3. The attempt at a solution

    i started by finding the two vectors
    rAC = -3,2 magnitued is sqrt(13)
    rBC = -2,-2 magnitued is sqrt(8)
    then i jumped into Enet = EAC + EBC
    did kq / r^3 * R for point AC
    k(-4e-6)/((sqrt(13))^3) * <-3,2> this gives an x and a y
    then the same for BC
    k(12e-6)/((sqrt(8))^3) * <-2,-2> this gives an x and a y

    i added x components togeather then the y components and said that was by
    Enet vector. then just found the magnitued.
    my Enet vector was some crazy numbers of
    that is not right lol
    and this is where im stuck
    any help would be great!!!
  2. jcsd
  3. Sep 2, 2009 #2


    User Avatar

    Re: Electric field???

    Check your arithmetic again, especially for the [tex]\vec j[/tex] component.
  4. Sep 2, 2009 #3
    Re: Electric field???

    Yeah, I did the problem, and your equations are correct (but I got a different answer!). You probably mistyped on your calculator, or added the components incorrectly.
    Last edited: Sep 2, 2009
  5. Sep 2, 2009 #4
    Re: Electric field???

    ok... here is what i did.
    first i started with k(-4e-6)/((sqrt(13))^3) * <-3,2> equation
    k = 9e9
    so 9e9*-4e-6 = -36000
    (sqrt(13)) ^ 3 = 46.8722
    so -36000/ 46.8722 = -768.04643
    then take that number and * by <-3,2> to get
    <2304.1393,-1536.09285> <x,y> respectivly
    do the same thing for k(12e-6)/((sqrt(8))^3) * <-2,-2>
    9e9 * 12e-6 = 108000
    (sqrt(8))^3 = 22.6274
    so 108000/22.6274 = 4772.970773
    then * <-2,-2> to get
    <-9545.94155,-9545.94155> <x,y> respectivly
    at the two x and the two y's and i get the same answer as posted??

    can you guys see where i went wrong?
    might be using the wrong k value?
    let me know

    well... i do get a different J component and it is -11082.0344 not what i had in the original post but my I component i get the same.
    is this right??
  6. Sep 2, 2009 #5
    Re: Electric field???

    never mind...
    i got it

    thanks everyone for the help!!!!
    just a stupid plug-in error lol
  7. Sep 3, 2009 #6
    Re: Electric field???

    1.a Cos a = 4/*65, E1 = 12/13k, E2 = 4/5k, use cos function theorem to find electric field!
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