Sphere a is attached to the ceiling of an elevator by a string. a second sphere is attached to the first one by a second string. both strings are of negligible mass. here m1 = m2 = m = 3.14 kg.(a) the elevator starts from rest and accelerates downward with a = 1.35 m/s2. what are the tensions in the two strings in newtons? t1 = nt2 =.(b) if the elevator moves upward instead with the same acceleration what will be the tension in the two strings in newtons? t1 = = .(c) the maximum tension the two strings can withstand is 92.6 n. what maximum upward acceleration (in m/s2) can the elevator have without having one of the strings /s2.

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