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Determine the electric field at the centre of the triangle

  1. Aug 10, 2012 #1
    physics6.jpg
    Consider an equilateral triangle of side 15.6cm. A charge of +2.0μC is placed at one vertex and charges -4μC each are placed at the two. Determine the electric field at the centre of the triangle.

    I used Pythagoreom theory to find the height and divided it by 2 but now I'm stuck because I realized it's not exactly the middle of the triangle, but rather the middle of the triangle side =/ How do I go from there to find the centre?

    Then I basically will use the formula E = kQ / r2, for the charge +2.0μC towards the centre. And correct me if I'm wrong, but do the electric field at the bottom end up cancelling out, or just the x-components of their electric field?
     
  2. jcsd
  3. Aug 10, 2012 #2
    See the dashed-line triangles? Can you see that their height + their dashed-line side = the height you have found?
     
  4. Aug 10, 2012 #3
    I'm sorry, I don't exactly follow. =/ Are you saying I don't need to do the Pythagorean process?
     
  5. Aug 10, 2012 #4
    You need to determine the position of the center. You have found the height of the triangle, but you are not sure how the height corresponds to the position of center. If you consider those dashed-line triangles, you will see how the center divides the height. Yes, you will need Pythagoras' theorem for that.
     
  6. Aug 10, 2012 #5
    mhmm, so here is what I did... With the 15.6cm, I used Pythagorean to find the height to be 13.5cm of the side triangle. Then Pythagorean theorom again to find the height of the center, which I calculated to be 11.0cm. Divided that by 2 to get the centre of the triangle.

    Am I in the right direction this time?
     
  7. Aug 10, 2012 #6
    Let's say the height of the bug triangle is H - this is what you have found.

    Let's say that the height of the "dashed" triangle (say, the one at the bottom) is h. Observe that (H - h) is equal to the length hypotenuse of one half of the "dashed" triangle. Can you relate H and h given all this, and find out what h is?
     
  8. Aug 12, 2012 #7
    I'm sorry, I'm really horrible visualizing this, but I think I get what you mean!

    So, I found the height of the bottom dashed triangle.
    I divided it in half...
    Used it with the height of the side triangle I originally found
    And with Pythagorean Theorom, found the height at the centre then divided it by 2 .... ?
     
  9. Aug 13, 2012 #8
    I am not sure I understand the reason for the final division. Can you show your intermediate results?
     
  10. Aug 14, 2012 #9
    So with the side triangle: I found the height to be 13.51cm
    With the bottom dashed triangle... because I need the height to reach the centre, I divided the height of that dashed triangle by 2, so I got: 6.7cm

    Using the side triangle height, the half height of the bottom dashed triangle and Pythagorean, I found the full height at the centre of the triangle to be: 11.7cm

    But because I only need the height from the tip of the equilateral triangle to the centre, I divided the full height by 2 and got: 5.9cm
     
  11. Aug 14, 2012 #10
    Let's label the vertices of the big triangle A (left), B (top), C (right). Let's call the point in the middle D. Let E denote the point of the intersection of line BD with line AC.

    You have found that the length of segment AE = H = 13.51 cm. That is correct.

    Now you need to find the length of segment DE = h = ?. To find this, you may consider triangle ADE. In this triangle, you know the length of AE, and you know what angle DAE is (what is it?). From these, you can find the length of DE.
     
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