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Homework Help: Equipotential surfaces electric field problem

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A given system has the equipotential surfaces shown in the figure
    What is the magnitude of the electric field?
    What is the direction of the electric field? (degrees from + x axis
    What is the shortest distance one can move to undergo a change in potential of 5.00

    2. Relevant equations
    E= V/d

    3. The attempt at a solution
    I seem to be getting the same answer and its wrong I used the pythagoreon therom to find distances of each and divided the volts by the respecitive distances. I do not know what I am doing wrong I keep getting 2.24*10^2 V/m. Moreover, when I do the angle part I just do some trig and use that. Either way I try to do it and get 26.6 or 63.2. What am I doing wrong? Is the first part just the slope
  2. jcsd
  3. Jan 27, 2010 #2


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    Where is the figure?
  4. Jan 27, 2010 #3
    I can't see the figure yet either, but unless you're dealing with a uniform electric field then you can't use the equation you listed. As you said, electric field is the slope of a potential vs. position graph. If you're using calculus, it's the derivative with respect to position.
  5. Jan 27, 2010 #4
    I think I uploaded it now. Sorry!
    It is an attachment

    Attached Files:

  6. Jan 27, 2010 #5


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    Slope of the equipotential lines is tan(theta) = 2/4 = 1.2.
    Find theta.
    Distance between the equipotential lines d is x*sin(theta). Here x = 4 cm.
  7. Jan 28, 2010 #6
    SO part A is the slope = -1/2? But when I do tan^-1 (2/4) I still get 26.6. What am I doing wrong?
  8. Jan 28, 2010 #7


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    What is the relationship between the directions of the electric field and the equipotential lines?
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