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Electric potential

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Two points (A and B) are shown in a constant electric field of E = 850 N/C. The distance between A and B is L = 2 m. The line joining the two points makes an angle of 40 degrees with the electric field. Determine the electric potential difference (in Volts) between points A and B -- that is VA - VB.


    2. Relevant equations

    E_s cos(theta)=-Ed


    3. The attempt at a solution

    would I just do E(cos theta) or -Ed? or are neither right?
     
  2. jcsd
  3. Sep 30, 2009 #2

    Delphi51

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    The way I like to think of it is that the change in potential V is the work that must be done per unit charge to move a charge from A to B. The work is the force that must be exerted times the distance for which it is exerted. So we have
    V = W/q and W = Fd or V = Fd/q
    The electric field causes a force on the charge we are moving, F = qE.
    However this force is not in the exact direction we are pushing the charge. So the force we need to overcome is actually qE*cos(A) where A is the angle between the E field and the direction we are going. That leaves us with a potential difference of
    V = Fd/q = qE*cos(A)*d/q.
     
  4. Sep 30, 2009 #3
    except that q is not given in the problem
     
  5. Sep 30, 2009 #4

    Delphi51

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    The q on the top cancels with the q on the bottom.
     
  6. Sep 30, 2009 #5
    yeah I thought that might be it so I tried it that way but still got the wrong answer
     
  7. Oct 1, 2009 #6

    Delphi51

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    I'm not supposed to show you the calculation for fear of spoiling the experience for you. But if you show your calc, we can check it.
     
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