Find Enet at center of equilateral triangel
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hi mate ur answer...
An equilateral triangle has three congruent sides, and is also an equiangular triangle with three congruent angles that each meansure 60 degrees. To find the height we divide the triangle into two special 30 - 60 - 90 right triangles by drawing a line from one corner to the center of the opposite side.
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Answer:
hey mate
here is your answer
Ecenter = E1 + (2)E2 = ? First calculate midpoint of triangle by splitting it down the middle, resulting in the top left angle being 30 degrees. To determine the height of the triangle I used cos30L where L = 0.156 m = 0.135 m. So midpoint will be half this = 0.0657 m. Now my first question is the E from Q1 going to simply be E = k2.0 x 10^-6/ 0.135^2? Or because Q1 (top of triangle) acts at angles towards Q2 and Q3 the E field felt from Q1 on p (middle) will be 2Ey after cancelling Ex due to symmetry? If going with my first solution we now have Ec = 9.876 x 10^5 N/C + 2Ey = ? Distance from Q2/Q3 to center = square of (0.078^2 + 0.0675^2) = 0.103 m. E = k(-4.0 x 10^-6)/0.103^2 = -3.39 x 10^-6. So Ey = sin30E = -1.696 x 10^6. Net E middle of triangle = E1 + 2Ey = 9.876 x 10^5 + (2)1.696 x 10^6 = 4.379 x 10^6 N/C? Unless the top charge has to be broken into components I think this is right?
thanks.