Question

3 charges lie on the x-axis. The positive charge q1=13uC is at the origin. The positive charge q3= 16UC is at 4.0m. Where must the negative charge q3 be placed so that the resultant electric force is zero?

Answer 1

arro x=0 x=d x=4m to make the V must be placed betwees q, and a net force pero The net force on 92 s 7 = 9, 9₂ (?) + 49 (?) 41&o da 4TE (4-d)? o = -9, 92 + V2 V3 Ad) 2 9, 92 - V2 V3 da Ad) 2 13x106 = 16x105 da (6-d) 2 13 (4-d)3 = 16 d2 13 (167d-8d) =16d2 208 +13d? - 1040 =16d² 30? +104d-208 = 0 solving the quadratic equatron d = 1.896m, -36.56 m. d can not be negative. Therefore the posntron is x = 1.896 m.