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Homework Help: Determin the Electric Potential Difference

  1. Sep 9, 2010 #1
    1. The problem statement, all variables and given/known data
    The drawing shows a uniform electric field that points in the negative y direction; the magnitude of the field is 3000 N/C. Determine the electric potential difference (a) VB - VA between points A and B, (b) VC - VB between points B and C, and (c) VA - VC between points C and A.


    2. Relevant equations

    3. The attempt at a solution
    I was able to solve the first two parts of this problem easily but I am hung up on how the angle of the triangle plays into this.
    I know A is zero because they are on the same equipotential field.
    B I solved by converting the distance to meters and multiplying that by the magnitude of the electric field.
    For C I know that it will be a combination of both the vertical and horizontal direction and I know that I am supposed to find that by using sin or cos but I just cant seem to put it together!
  2. jcsd
  3. Sep 9, 2010 #2


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    Homework Helper

    Looks good so far, Kvarner.
    At the risk of boring you, I suggest going back to the definition of electric potential as the work per charge needed to move a charge through an electric field. When moving the charge horizontally from A to B, there is no force in the direction of movement so no work need be done. Like moving horizontally in a gravitational field. So, moving from A to C you could do A to B first with work, energy and potential difference zero. Then add the potential difference for the B to C move that you've already calculated. Hope this makes sense ...

    Alternatively, going straight from A to B the electric force is purely vertical so the pd is the work/charge = F*d/q = q*E*d/q where only the vertical component of distance is used because that is the direction of the force. I suppose it should really be written as a dot product F•d.
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