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 charges

  1. Mar 1, 2007 #1
    Two electric charges, q1 = +22.8 nC and q2 = +10.1 nC, are located on the x−axis at x = 0 m and x = 1.00 m, respectively. What is the magnitude of the electric field at the point x = 0.50 m, y = 0.50 m?

    I used the equation

    E= ((k)(q))/r^2
    for each q1 and q2.

    I don't know what to do next to find the mag. at x = 0.5 and y = 0.5.

    thanks in advance
  2. jcsd
  3. Mar 1, 2007 #2
    try finding the E at (0.50,0.50) from each charge using the equation.
    then it's a matter of summing both electric fields.(Vector algebra).
  4. Mar 1, 2007 #3
    to find magnitude: think Pythagorean ;)
  5. Mar 1, 2007 #4
    when you say use 0.5, 0.5... does that mean that r = 0.5?
    so it would be 0.5^2 for the denominator?
  6. Mar 1, 2007 #5
    find the distance from the charge to the points(0.5,0.5) (this is the xy coordinate for the point in space where you are trying to find the electric field)
  7. Mar 1, 2007 #6
    Sorry I'm not understanding it.

    There is a distance of 1.0 m between q1 and q2.
    The points (0.5, 0.5) would be in between q1 and q2?
    So the distance would be 0.5m and thus r = 0.5?
  8. Mar 1, 2007 #7
    you are picturing it wrong. when you draw the picture you have q1 at (0,0)ie, x=0 and y=0 for q1. similarily, for q2 it is (1,0) where x=1 and y=0. Now you are required to find the the electric field at (0.5,0.5) that is x=0.5 and y=0.5. Hope you have a picture now. when you are finding the E from q1 at (0.5,0.5) , the distance is (0.5-0,0.5-0) = (0.5,0.5) therefore r = sqrt(0.5^2+0.5^2). use this 'r' value into E equation, that is your E at (0.5,0.5) from q1.Do the same for q2 too. Then follow what i mentioned above
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electric charges
  1. Electric Charges (Replies: 1)

  2. Electric charge (Replies: 3)