Physics, Two charges separated by one meter exert 1-N forces on each other. If the charges are pushed...

Two charges separated by one meter exert 1-N forces on each other. If the charges are pushed to 1/4 meter separation, the force on each charge will be
A) 1 N.
B) 2 N.
C) 4 N.
D) 8 N.
E) 16 N.

Please, explain in detail

Thank you

You must use Coulomb's law (I think that's the name of it, I may well have misspelled it) which is
K Q1*Q2/(d^2) k being the electric constant (8.99*10^9) Q1 being the charge on one object, Q2 the charge on the second and d the distance between the charges. Since the force is proportional to the inverse square of the distance, quartering the distance will multiply the force by a factor of 16 (because 1/((1/4)^2)=1/(1/16)=16

F = kqQ/r²

Let r' = 0.25r

F' = kqQ/r'² = 16kqQ/(0.25r)² = 16F

Since F = 1N, we have

F' = 16N

Yin