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Changing the direction of the acceleration Vector

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    2 point charges. q1 is -25 microC and q2 is 50 microC are separated by a distance of .1m where q1 is on a line that connects q1 to q2 and q1 is to the left of q2. point p is .02 m to the right of q1 and .8m to the left of q2.


    2. Relevant equations

    E = F/q

    E(e) = Force on electron due to Electric Field

    3. The attempt at a solution

    okay, so the electric field at point p due to these 2 point charges is -6.3 E 8 N/C dot i.

    if we put an electron at point p, the electron will feel an acceleration.

    the electric field points to the left at point p. yet, the electron feels an acceleration vector pointing to the right. How do I switch the direction of the acceleration vector without multiplying a negative to the equation E(-e) = m(acc) where E = -6.3 E 8 N/C dot i ?????
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 14, 2009 #2
    so? anybody with any ideas?
     
  4. Feb 14, 2009 #3

    LowlyPion

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    Homework Helper

    Your Force equation is the product of the charge scalar and the E-field. Your e-Field at p is -x. But there is a - carried with the e. This reverses the acceleration that the charge will experience.
     
  5. Feb 14, 2009 #4
    there's got to be another way to switch the direction without multiplying by the negative scalar. If you look inside the Force equation...

    f = q q k / r squared dot unit vector r. however, i'm not sure if the positive test charge feels a negative or positive force from the source.
     
  6. Feb 14, 2009 #5

    LowlyPion

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    Positive test charge experiences positive force (repulsive force) outward from the point charge. The r vector of force is negative directed at that point as a result of the negative position from the net outward from the dominating positive. As a negative the electron however wants to accelerate opposite, i.e. toward the positive point, against the positive gradient, but of course only there and not beyond.
     
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