A charge -q and a charge 4/9q are located at coordinates x = 0 and x = a, respectively. If you place a third charge so that all three are in static equilibrium, justify that its position and its charge,respectively, can only be
The Attempt at a Solution
Since the three charges are in static equilibrium I know that the electric field is zero. Which means each individual electric field has to add up to zero as a whole. If F_ab = (q_1)(q_2)/r^2 is get F_ab= (-4q^2)/9a^2, so does that mean the third charge added in has to cancel that out?