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B Electric field direction from high V to lower V

  1. Oct 24, 2015 #1
    http://postimg.org/image/we3bnkyqn/
    I know electric field points in direction in which the potential is decreasing.
    How does potential decrease in that direction?
    Here is what I think
    potential at point a is 10v and at point b it is 20v
    and direction of E points from b to a,hence
    electric field points in direction in which the potential is decreasing.
    k.png
    Am I thinking right?
     
  2. jcsd
  3. Oct 24, 2015 #2

    sophiecentaur

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    The Field is at right angles to the Equipotential surfaces.
     
  4. Oct 25, 2015 #3
    I would add this to what sophiecentaur says:
    I am quoting the first two lines of the original question:
    "I know electric field points in direction in which the potential is decreasing.
    How does potential decrease in that direction?"
    It is not that there is some other reason for potential to decrease in that direction. The direction of the electric field and the direction of decrease of potential energy are the same. If there is an electric field pointing in a certain direction, the potential decreases in that direction. Equivalently, if the potential decreases in a particular direction, the electric field points in that direction
     
  5. Oct 25, 2015 #4

    Mister T

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    Gravitational potential energy decreases when a ball of mass m is lowered from the ceiling to the floor. Note that it moves in the direction of the gravitational field ##\vec{g}##.

    Likewise, electric potential energy decreases when a particle of positive charge q moves in the direction of the electric field ##\vec{E}##. Since electric potential is simply electric potential energy divided by q, it also decreases.

    Now, things get scrambled up a bit when the particle has a negative charge, because then the potential energy would increase, but the electric potential would still decrease.
     
  6. Oct 25, 2015 #5

    Mister T

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    Not quite. Imagine a particle of negligible mass and positive charge located at Point b. Release it and it wouldn't move towards Point a. It would instead move towards the 10-volt line along a path of shorter distance. The electric field ##\vec{E}## is tangent to that path at each point.
     
  7. Oct 26, 2015 #6
    Yes.That's what I have drawn.
     
  8. Oct 26, 2015 #7
    No,according to my book
     
  9. Oct 26, 2015 #8

    haruspex

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    But it doesn't look like that. If your diagram is simply 2D, your field lines are at about 60 degrees to the equipotentials. Were you attempting to show a 3D view in perspective?
     
  10. Oct 26, 2015 #9

    ehild

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    That is not enough. Imagine you are on a hill. You can go down and decrease your potential in lot of ways. You can choose a tourist way descending slowly. Or you can go straight down along the steepest direction. What direction would a water flow choose? It is analogous with the charges. They accelerate along the electric field. So what is the direction of the electric field?
    That is wrong.
    No, your logic is wrong.
     
    Last edited: Oct 26, 2015
  11. Oct 26, 2015 #10
    yes,sort of.
     
  12. Oct 26, 2015 #11

    ehild

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    Then show all the three axes, x, y, z, and equipotential surfaces instead of lines.
     
  13. Oct 26, 2015 #12
    Actually ,I don't know .In my book it was an example ("solved problem)and it has been drawn in the same manner as I posted.I wanted to know the reason.
     
  14. Oct 26, 2015 #13

    ehild

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    And what was the text to the problem? Copy it fully, please.
     
  15. Oct 26, 2015 #14
    This was the question.some equipotential surfaces are shown in the figure below What can you say about the magnitude and the direction of the electric field?http://postimg.org/image/k6cm9b9wf/
     
  16. Oct 26, 2015 #15

    ehild

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    Well, we can imagine that the equipotential surfaces are planes and the drawing shows the projections on the (xy) plane, so the surfaces are vertical (out of the page). There are no a, b points and E vector. You see distances along the x axis. You see the angle, that the lines enclose with the x axis. And you see the potential value along the equipotentials.
    What did you learn, how is the electric field and the potential related?
     
  17. Oct 26, 2015 #16
    No,there are not any a and b but there is E vector present in the solution part,what I wrote in post #14 was just a question.
     
  18. Oct 26, 2015 #17

    ehild

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    What is the potential difference between the equipotential surfaces shown?
    Remember how the potential difference was defined with work.
     
  19. Oct 26, 2015 #18
    The part of a solution says Electric field lines are always perpendicular to the local equipotential surface .electric field points in direction in which the potential is decreasing And then the below image was given.http://postimg.org/image/we3bnkyqn/
     
  20. Oct 26, 2015 #19
    10 V.
     
  21. Oct 26, 2015 #20
    The part of a solution says Electric field lines are always perpendicular to the local equipotential surface .electric field points in direction in which the potential is decreasing.
    ##W##=##q####Δv##
     
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