Two charges q1 = -25 and q3 = 4.0 are placed 0.012 m apart. A third negative charge q2 is placed a distance L from q1 as shown. If the net electric force on q2 is 0 N, what is the value of L?

Ohm's law tells you that i = v/r where v is equivalent voltage and r is equivalent resistance of the circuit
since all circuits have the same voltage, your question is really asking which circuit has the lowest total resistance, since the lower the resistance, the higher the current
circuit a has a resistance of 5 ohms; circuit c is in series so has a resistance of 10 ohms; circuit b is in parallel and therefore has a resistance of 2.5 ohms (review your notes on adding resistances in parallel and series); therefore b has the highest current

Caused conformity rates to go down. apex ; )

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