Two small speakers, 0.650 m apart, are facing in the same direction. They are driven by one 677 Hz oscillator and therefore emit identical sound waves in phase with one another at the respective points of origin. (The speed of sound waves in air is 343 m/s.) (A) An observer wishes to stand in front of one of the speakers, at the closest point (i. e., smallest x-value) where intensity is at a relative maximum. At what distance x from the nearest speaker should she position herself? (Enter your answer in m.)
(B) The observer now wishes to stand at the closest point along that line where intensity is at a relative minimum. At what distance x should she position herself now? (Enter your answer in m.)

Ok what do you need with

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge