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Homework Help: Point charges and magnitude of the electric field at the center of a triangle

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A point charge q = 4.9 \mu C is placed at each corner of an equilateral triangle with sides 0.20 m in length.
    Is the magnitude of the electric field at the center of the triangle greater than, less than, or the same as the magnitude at the midpoint of a side? Explain.

    2. Relevant equations
    E = q/r^2

    3. The attempt at a solution
    I found the magnitude of the midpoint of the triangle to be 1.5e6 but I am not sure how to explain this in words. Any guidance would be appreciated!
  2. jcsd
  3. Jan 27, 2010 #2
    well you know that the magnitude of the electric field at the centre of the triangle is 0.
    and you also know that that if you have equal charges placed a distance d apart on the line, at d/2 the E field is 0. But in this case there is a third charge on the opposing vertex. So the electric field there cannot be 0. you also know that they are all positive charges. So there you go
  4. Jan 27, 2010 #3
    oh sorry I thought you said you got the magnitude on the side to be 1.5e6.
    Well keep trying it until you get 0 at the centre lol
  5. Jan 27, 2010 #4
    oops i meant to say the midpoint of the sides was 1.5e6. sorry!
  6. Jan 27, 2010 #5
    so to put it in stupid people (being me) terms, since we know that the first two point charges (4.9 uC) are positive, and i calculated the third pt charge to be positive, this means that the midpoint of the triangle is zero? so the zero mag is due to the repelling forces among the 3?
  7. Jan 27, 2010 #6
    Yeah, its a property of certain shapes. I think it has something to do with all shapes that have as many symmetrical vertex to midpoint bisects as they do sides all will experience the same 0 force at the centre. The same thing is for gravity right?

    Also, thats why circles are ridiculous, because they have infinite sides and infinite symmetrical lines can be drawn through it.
  8. Jan 27, 2010 #7
    ok thanks
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