This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise The three charged particles in the figure below are at the vertices of an isosceles triangle (where d = 3.90 cm). Taking q = 6.50 JC, calculate the electric potential at point A, the midpoint of the base. 2d - 91 Part 1 of 3 - Conceptualize Since the midpoint of the base is closer to the two negative charges, it probably has a negative potential on the order of several kilovolts. Part 2 of 3 - Categorize Each charge creates a potential at the midpoint of the base. In order to find the total electric potential at this point, we add the contributions from each of the charges. Part 3 of 3 - Analyze The charged particle at the top vertex of the triangle is a distance /1 from point A. By the Pythagorean theorem, we have ( 20)2 = ( # ) + 512 . Solving for the unknown distance, we obtain the following. 1 = \\/ 402 - 12 V15 d = 0.0755 m The distance /2 of the charge to the left and the distance ry of the charge to the right from point A are /2 = /3 = d/2. The potential at point A is the sum of the contributions from each charge at its distance from point A as follows. V = Ke( 19 - 72 - 8.99 x 109 N . m2/C2) 6.50 10 * 10- 6 c ) [ X Your response differs from the correct answer by more than 100%. /m) * 106 V