## rock_mit182 one year ago 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 }

1. anonymous

2. rock_mit182

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

3. rock_mit182

0,1 meters for the initial values

4. rock_mit182

now if the radius is multiplied by 1/2

5. anonymous

yes

6. rock_mit182

the radius would be 1/20, right ?

7. anonymous

yea you got it keep going

8. rock_mit182

im stuck

9. rock_mit182

i think the right options are 8N or 16 N

10. anonymous

16 N

11. rock_mit182

because r is squared, right ?

12. anonymous

yea

13. rock_mit182

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 ?

14. rock_mit182

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

15. anonymous

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

16. anonymous

you get it?

17. rock_mit182

what did you do ?

18. rock_mit182

did you see my last equation ?

19. rock_mit182

i curious about what just did happen

20. anonymous

which one?

21. rock_mit182

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

22. rock_mit182

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

23. rock_mit182

im sorry F1 =4 N

24. anonymous

oka what is the question asking? id dont see

25. rock_mit182

F2 is incresing respect to one by a factor of ?

26. anonymous

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

27. rock_mit182

and i think is by a factor of 4

28. rock_mit182

anyway thanks a lot for helping me out

29. anonymous

no problem anytime

30. rock_mit182

@IrishBoy123 any thoughts ?

31. IrishBoy123

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]

32. rock_mit182

yeah that is what i meant

33. rock_mit182

distance is halved

34. IrishBoy123

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