1. Not finding help here? Sign up for a free 30min 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!

Electric forces and fields

  1. Nov 5, 2004 #1
    Two 2.06e-6 C point charges are located on the x axis. One is at 1.06 M and the other is at -1.06 M

    (a) Determine the electric field on the y axis at .48 M

    This is how i did it:

    Etotal=sum of E vectors...

    E=ke*q/r^2

    r^2=1.06^2+.48^2

    q=2.06e-6

    E1 = Ke*q/(1.06^2+.48^2)
    E2= same thing

    E1+E2 = total..
    answer is wrong..what am i doing wrong
     
  2. jcsd
  3. Nov 5, 2004 #2
    Both charges are positive. The electric field equation is given by

    Kq/r^2 * r-hat, which is a unit vector pointing radially away from the charge.

    If you draw a picture of the two e-fields, you'll see that the components don't add exactly (in fact, the x components cancel out), so you need to take that into account.
     
  4. Nov 5, 2004 #3
    yeah when i was trying to do this i just realized that too..but i still dont nkow how to find the electric field
     
  5. Nov 5, 2004 #4
    You should ask yourself: what is the unit vector, r-hat, that points from the point charge to the coordinate in consideration?

    IE, for one point charge, what is the unit vector that points from (1.06, 0) to (0, 0.48)? Knowing this, you can express the unit vector r, in terms of cartesian coordinates, and you should see, that when you add the E-fields together, the x-hat components cancel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electric forces and fields
Loading...