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Electric Field Intensity

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Find Enet (magnitude and direction) at point A and include a vector diagram.

    q1 = 2.0x10^-6 C
    q2 = 6.0x10^-6 C
    r1 = 2.5x10^-3 m
    r2 = 1.7x10^-3 m

    2. Relevant equations
    E = Klql/r^2 or E = Kq/r^2 (the first equation just has absolute values around the q for reminder but they both are the same thing)

    3. The attempt at a solution
    First I looked at the diagram and I know that the electric field created by a charge will always put away from a positive charge and towards a negative so I drew a vector diagram for that at point "A" using tip to tail method, then I joined E1(electric field created by q1) and E2(electric field created by q2) to have a resultant of E.

    I then used the equation E = kq/r^2 to find out what E1 and E2 are and I found out that they equal 2.8x10^9 and 1.9x10^10 respectively. My question, now I have to use the component method of vector addition to find out what E is but I need an angle to do that. Is the angle in this case 90 degrees?

    Attached Files:

  2. jcsd
  3. Dec 16, 2008 #2

    Doc Al

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    Staff: Mentor

    How did you find E2? What value of r (or r^2) did you use?
    No. Figure out the angle that E2 makes with the horizontal by examining that triangle and using a little trig.
  4. Dec 16, 2008 #3
    I figured out E2 by using the values q2 = 6.0x10^-6C and r(or r^2) = 1.7x10^-3m. Because these values were given at the start of the question, we didn't need to find them.

    Then I took those numbers and I directly plugged it into the equation E= kq/r^2.
    Last edited by a moderator: Dec 16, 2008
  5. Dec 16, 2008 #4

    Doc Al

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    Staff: Mentor

    To figure out E2 (the field from q2 at point A) you need the distance from q2 to point A, not the distance between the charges (which is given as 1.7x10^-3m).
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