y/SUB]1. The problem statement, all variables and given/known data Determine the electric field at the center of the triangle. All sides are 15.6 cm. Vetex is 2.0 uC (Q1) while the bottom points are both -4.0 uC (Q2 and Q3). 2. Relevant equations E = kQ/r^2 E = F/q 3. The attempt at a solution 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.