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Homework Help: Equipotential surfaces

  1. Sep 8, 2006 #1
    Problem:

    A given system has the equipotential surfaces shown below, where Vo = 12.0 V.

    20-26alt.gif

    (a) What are the magnitude and direction of the electric field?
    (b) What is the shortest distance one can move to undergo a change in potential of 5.00 V?

    I am not too sure on what equations I can use to solve this, but I have a feeling one of them will be delta V = -W/qo, and I have no other ideas on how to approach this problem.
     
  2. jcsd
  3. Sep 9, 2006 #2
    You need trig to figure out the distance between the equipotentials. The electric field is always going to be orthogonal to equipotential surfaces. Do you know the relationship between E, V and r? Use that to find E, then use the E that you got from part a) for the last part.
     
  4. Sep 9, 2006 #3
    Okay, so for part (a), I use E = - V/r = - 12/4 = -3? Did I interpret your post correctly?
     
  5. Sep 9, 2006 #4
    Not quite, you need the distance between two equipotential lines, which would be along a line perpendicular to them. I think you've gone along the x axis, which is at an angle to the red lines. Use the triangle formed by the x and y axes and the line V=V0, split it into two smaller right angled triangles and look for similar angles.
     
  6. Sep 9, 2006 #5
    Okay, I determined how to split the triangle into two smaller right triangles and found the smiliar angles, but what exactly is this telling me?
     
  7. Sep 9, 2006 #6
    You should be able to get the distance between the equipotentials, which is your r in E=-V/r. Also, one of the angles gives you the direction of the E field.
     
  8. Sep 9, 2006 #7
    Oh, okay, so the distance between the lines is 1.78885 m, so this is my r. So plugging this into E = -V/r = -12/1.78885 = -6.7082 V/m?

    Edit: I just tried submitting this answer (both negative and positive), and both were incorrect.
     
    Last edited: Sep 9, 2006
  9. Sep 10, 2006 #8
    That's odd, that's the answer I got. Have you added the direction? Does it say how many s.f. to give?
     
  10. Sep 10, 2006 #9
    I found my mistake. The units in the chart are in cm, so I converted all values into m. This gave me a value of 670.82, which is a correct answer. I also calculated the correct answer for the direction, which is 243 degrees counterclockwise from the +x axis. I also solved part (b). Thank you for your help!
     
    Last edited: Sep 10, 2006
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