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About electric fields (Symmetry)

  1. Mar 15, 2017 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    1) Imagine you have a clock but in each number you have a charge "q". If I have a test charge Q at the center, then by simetry, the net force on Q is zero. I imagined like, 1 cancels 7, 2 cancels 8, etc...
    But then, if I have a polygon with 13 sides, with a test charge Q, the net force also would be zero!
    How come?
    (This is the very first exercise in Griffiths book, 2.1)

    2) Also, that very commom problem of two charges, apart from a distance "d" when I need to find the Eletric Field in a midpoint between these two chages placed on z axis.
    If the charges are +q and +q, I only consider the vertical component of the eletric field, but if the charges are +q and -q then I only consider the horizontal component. Why??
    I don't understand why!

    I don't need help with the math per se, but I'm missing something about understanding simetry in Electric Fields...
    Thanks in advance for an explanation!
     
  2. jcsd
  3. Mar 15, 2017 #2

    blue_leaf77

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    1) For the sake of simplicity imagine you have an equilateral triangle with identical charges placed at each corner and a test charge in the center. Assume that the net force at the center is nonzero and directed in certain direction. Now rotate the whole charge arrangement by 1200 and 2400, you will get the same system. Since you rotate the system, the net force at the center must also follow the rotation right? How can you reconcile the fact that the system stays unchanged but the field is changed?

    2) It's not clear in which way the charges are arranged. Please describe in terms of axis, not just vertical and horizontal.
     
  4. Mar 15, 2017 #3

    robphy

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    2) At a point of interest, draw/sketch the electric field vector there due to each charge... then graphically add them.
     
  5. Mar 16, 2017 #4
    I'm sorry. I made got this drawn below.
    In "a" both charges are the same signal (positive) and in "b" one is positive and the other is negative. As shown in the figure, I don't understand why the total field is the way it is...
    Like, in "a" we have "2q cosθ" because the said horizontal components cancel.
    In "b" we have "2q sinθ" because the said vertical components cancel.
    I don't even understand why in she second case "b" the Eletric Field points to where it is pointing...

    ***Never mind...I think I got it now. I was trying to do all in my mind but when I drawn I got it.
    Sorry everyone!!
    z7M6TUp.jpg
     
  6. Mar 16, 2017 #5

    robphy

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    Electric fields satisfy the Superposition principle....
    which means the total electric field vector at a point in space
    is equal to the vector-sum of the electric field vectors at that point (from each source taken one at a time)

    Can you add the vectors graphically (parallelogram rule)?
     
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