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Potential difference

  1. Jul 7, 2008 #1
    Just wondering how do you calculate the potential difference between Pt A and B in the picture.

    [​IMG]

    Value of E: 95 N/C
    AB=0.33 m
    the angle is 42

    I researched and found an equation V= Ercostheta
    but I've never seen or used this equation before. Will I get the potential difference between A and B just by pluggin the values in?
    V=(95)(0.33)cos(42)
     
  2. jcsd
  3. Jul 7, 2008 #2

    cepheid

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    Staff Emeritus
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    Gold Member

    Two methods that lead to the same answer:

    1. Start with the definition: The potential difference between A and B is the line integral of the electric field along the path from A to B. However, since E is uniform (constant in magnitude and direction), the integral turns into something really simple.

    2. Avoid setting up the integral in the first place by just recognizing that electric potential is only increased when going *against* the field. That's why the vertical component of the path doesn't matter, and the voltage is just the horizontal component of the path multiplied by the electric field strength. It's because you're considering this horizontal component that the cosine comes in.
     
  4. Jul 7, 2008 #3
    thanks! :)
     
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