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Ranking E field in equipotential surfaces, confused!

  1. Sep 25, 2005 #1
    Hello everyone i'm confused on this topic. I read about it in the book and it made sense though. The question is: Figure 24-25 shows three sets of cross sections of equipotential surfaces; all three cover the same size region of space.
    Diagram: http://www.webassign.net/hrw/25_29.gif

    (a) Rank the arrangements according to the magnitude of the electric field present in the region, greatest first (use only the symbols > or =, for example 1=2>3).


    (b) In which is the electric field directed down the page?


    Well in the book it says, E is always perpendicular to the equipotential surfaces and I get that, also i understand that if a charged particle moves from one end to the other but still ends up on the same equipotential surface, the work done is 0. Also if you moved a charged particle to a difference surface, and u moved another charged particle to the same surface, the work done would be equal to eachother, because they are both on the same equipotential surface. But for part (a) i'm confused, what equation would I use to calculate the E field, do i assume there is a charged particle moving from each surface to the next until it hits the last one or what? Also for part (b), wouldn't they all be pointing downwards? if i assumed there was a charged particle star4ting at the top of the surfaces, because a + charges E field is always directing away from it, wouldn't they all be down the page?:bugeye:
     
  2. jcsd
  3. Sep 27, 2005 #2

    mukundpa

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    Homework Helper

    [tex] E = - dV/dX = - \Delta V/ \Delta x [/tex] for unidirectional field, gives that potential decrease in the direction field hence field is upward in 1 and 2 but downward in 3.

    for first
    [tex] \ E_1 = - (80 -100)/x = 20/x [/tex]
    for second
    [tex] \ E_2 = - (-120+100)/2x = 20/2x [/tex]
    for third
    [tex] \ E_3 = - (-30 + 50)/2x = - 20/2x [/tex]
    so 1>2=3
     
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