Please someone help to figure this out:
the problem says that the force between two spheres is 4N they're separeted by 10 cm. If the distance is two times less the initial what will happen with the magnitud of the force between them.
columb's laws = F =k\frac{ q*q }{ r^2 }

Question

Please someone help to figure this out: 
the problem says that the force between two spheres is 4N they're separeted by 10 cm. If the distance is two times less the initial what will happen with the magnitud of the force between them.
columb's laws = F =k\frac{   q*q   }{ r^2 }

anonymous · Answer

i can help you

rock_mit182 · Answer

$$4 = k \frac{ q*q }{r^2 } $$

rock_mit182 · Answer

0,1 meters for the initial values

rock_mit182 · Answer

now if the radius is multiplied by 1/2

anonymous · Answer

yes

rock_mit182 · Answer

the radius would be 1/20, right ?

anonymous · Answer

yea you got it keep going

rock_mit182 · Answer

im stuck

rock_mit182 · Answer

i think the right options are 8N or 16 N

anonymous · Answer

16 N

rock_mit182 · Answer

because  r is squared, right ?

anonymous · Answer

yea

rock_mit182 · Answer

now if de charge of the two spheres is two times each one, mantaining the initial radius an force, the answer would be the same ?

rock_mit182 · Answer

$$4 = k \frac{ 2q*2q }{ (0,1)^2 }$$

anonymous · Answer

okay so the initial force  between the two charges separated by a distance d is F = 2kQ2 / d2 .
After touching, the charges become Q/2 and 5Q/4 and the force is 5kQ2 / 8d2 = 5F / 16 . so the answer would be 5F / 16

anonymous · Answer

you get it?

rock_mit182 · Answer

what did you do  ?

rock_mit182 · Answer

did you see my last equation 
?

rock_mit182 · Answer

i curious about what just did happen

anonymous · Answer

which one?

rock_mit182 · Answer

$$4 =k \frac{ 2q *2q }{ (0,1)^2 }$$

rock_mit182 · Answer

F1 =2 so F2 = ? ; keeping in mind that Im  using the initial values but double charge ( each one)

rock_mit182 · Answer

im sorry F1 =4 N

anonymous · Answer

oka what is the question asking? id dont see

rock_mit182 · Answer

F2 is incresing respect to one by a factor of  ?

anonymous · Answer

is it multiple choice  i think its increasing by 2 but im not sure you might wanna ask someone else for a second opinion

rock_mit182 · Answer

and i think is by a factor of 4

rock_mit182 · Answer

anyway thanks a lot for helping me out

anonymous · Answer

no problem anytime

rock_mit182 · Answer

@IrishBoy123 any thoughts ?

irishboy123 · Answer

you have an inverse square relationship here so you can say $\large F = \frac{C}{r^2} $ where C is a constant

ie, $\large C = F \ r^2$

so, applying this:
 $\large C = 4 \times 10^2 = F \times 5^2$

what is F?

[PS "two times less the" doesn't mean very much so i assumed you meant distance is halved]

rock_mit182 · Answer

yeah that is what i meant

rock_mit182 · Answer

distance is halved

irishboy123 · Answer

so
$$4 \times 10^2 = ?? \times 5^2$$