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Homework Help: Electric Fields -- Charges at corners of equilateral triangle

  1. Jul 29, 2014 #1
    Consider an equilateral triangle of side 15.6 cm. A charge of +2.0uc is placed at one vertex and charges of -4.0C uc each are placed at the other two, as shown in the diagram to the right. Determine the electric field at the centre of the triangle

    ANgle= 60 sides--> d1= d2=d3=0.156m q1=2X10^-6C q2=q3=-4X10^-6C

    E=kQ/r^2 {E= E1+E2+E3} Ex= Ex1 +Ex2+Ex3 Ey= Ey1+Ey2+Ey3
    Ex.=(+-)E.CosAngle Ey.=(+-)E.Sin(angle)

    Let q1 be the charge on top, q2 bottom left, q3 bottom right. Side d1, left side, side d2 , right side and side d3, bottom side of triangle.

    3. The attempt at a solution
    I am having trouble witch direction the E components are going. That is for example, which direction is the Ex2 going. For example if its going towards the left then I would put a negative sign in front of E2 --> Ex2=-E2Cos(angle)

    Is it the following ? that E2x and E2y would be negative,(pointing out of the triangle) E3x would be positive,E3y would be negative,(also pointing out,towards charge 3) and E1x is N/A, and E1y is negative(poiting towards the middle )?
  2. jcsd
  3. Jul 29, 2014 #2


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    Staff: Mentor

    I suspect that you've got the right idea, but your descriptions of directions are a bit fuzzy. For example, any vector with an origin (tail) inside the triangle must be "pointing out of the triangle" regardless of which direction it actually points! If you follow along the direction of the arrow of the vector you must eventually cross to the outside.

    The best way to handle these sorts of problems is to make a sketch, drawing in the directions of the fields due to each charge at the point in question. Remember that positive charges have a field that radiates radially outward (from the charge), while negative charges have a field that radiates inwards (towards the charge). Once you've done that it's a simple matter to check that your component signs agree with the sketch.
  4. Jul 30, 2014 #3
    alright, umm I had already done a sketch-Im just not great at it l-lol

    So does that mean, that the direction of the field is always independent of the other charges. Like the fact that the forces of the negative charges would repel, and therefore there forces would point outwards in this case, doesn't affect the direction of where there fields point.Is that correct?

    If thats so, then I would assume the positive charge's field,E1 would point outwards, so there E1y would be positive. And the negative charges fields' would point towards the inside.or its charge, So Ex2 would be positive, Ex3 would be negative. And both Ey2 and Ey3 would be positive. Is that better?
  5. Jul 30, 2014 #4


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    Staff: Mentor

    The fields due to each charge are independent and simply add at any given point... the total field is given by the sum of the contributions due to the individual charges taken one at a time. In this sense, then, charges do not interfere with each other. The fact that like charges repel does not in any way change the fields that those charges individually produce at a given location.

    The direction of the field produced at a given location is directed along a line joining the (point) charge to that location. The field will be directed in the opposite direction of the charge if the charge is positive, and towards the charge if it is negative.

    I'm not sure why you keep referring to "pointing inside" or "pointing outside". The vectors you're looking for are not attached to the charges themselves. You are looking for the field vectors at the center point of the triangle: The origin of those vectors are "attached" to that point and will be directed either to or way from the associated charge causing that field.


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  6. Jul 30, 2014 #5


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    Staff: Mentor

    Have you looked at the 5 threads listed at the foot of this page? Right down at the very end.....
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