2 point charges q1 = 26 micro columbs q2 = -47 micro columbs what distance must

Question

2 point charges
q1 = 26 micro columbs
q2 = -47 micro columbs

what distance must they be apart to have electrostatic force of5.7N?

formula

F = k(q1)(q2) / r^2
k = 8.99*10^9 (N*m/C^2)

of course putting in q1,q2,k, and 5.7 for F..and solving for rgives me the answer 1.39 which is whats expected but WHY DO WE NOTHAVE TO divide out or convert from microC to just C?

why do i get correct answer without converting?
Answer is 1.39m (from back of book), but when you just plug in youget 1.39micrometers (which is not right..)

i will rate you to the max if you can tell me why

Explanation / Answer

q1 = 26 micro columbs
q2 = -47 micro columbs

Electrostatic force F = 5.7N

we know F = k(q1)(q2) / r^2
where k = coulomb's constant = 8.99*10^9 (N*m ^ 2 /C^2)
from above formula separation of the charges  r = [ k (q1)(q2) / F ]                                                                           = { [ 8.99*10^9*26*10^-6*47*10^-6 ] / 5.7 }                                                                           = [ 1.927]                                                                           = 1.388 m                                                                           ~ 1.39 m

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