There are four traffic lights on a stretch of road and the distance between each traffic light is 300 m. There is a 10-s lag time between successive green lights: The second light turns green 10 s after the first light turns green, the third light turns green 10 s after the second light does (and, thus, 20 s after the first light turns green), and so on. Each light stays green for 15 s. Having watched this sequence for some time, you understand that you can drive through the stretch of road at a range of constant speeds. You drive through the first light right whenit turned green. What is the fastest speed that you can drive through the stretch of road without running any red lights

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