rock_mit182
  • rock_mit182
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 }
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i can help you
rock_mit182
  • rock_mit182
\[4 = k \frac{ q*q }{r^2 } \]
rock_mit182
  • rock_mit182
0,1 meters for the initial values

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More answers

rock_mit182
  • rock_mit182
now if the radius is multiplied by 1/2
anonymous
  • anonymous
yes
rock_mit182
  • rock_mit182
the radius would be 1/20, right ?
anonymous
  • anonymous
yea you got it keep going
rock_mit182
  • rock_mit182
im stuck
rock_mit182
  • rock_mit182
i think the right options are 8N or 16 N
anonymous
  • anonymous
16 N
rock_mit182
  • rock_mit182
because r is squared, right ?
anonymous
  • anonymous
yea
rock_mit182
  • 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 ?
rock_mit182
  • rock_mit182
\[4 = k \frac{ 2q*2q }{ (0,1)^2 }\]
anonymous
  • 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
anonymous
  • anonymous
you get it?
rock_mit182
  • rock_mit182
what did you do ?
rock_mit182
  • rock_mit182
did you see my last equation ?
rock_mit182
  • rock_mit182
i curious about what just did happen
anonymous
  • anonymous
which one?
rock_mit182
  • rock_mit182
\[4 =k \frac{ 2q *2q }{ (0,1)^2 }\]
rock_mit182
  • rock_mit182
F1 =2 so F2 = ? ; keeping in mind that Im using the initial values but double charge ( each one)
rock_mit182
  • rock_mit182
im sorry F1 =4 N
anonymous
  • anonymous
oka what is the question asking? id dont see
rock_mit182
  • rock_mit182
F2 is incresing respect to one by a factor of ?
anonymous
  • 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
rock_mit182
  • rock_mit182
and i think is by a factor of 4
rock_mit182
  • rock_mit182
anyway thanks a lot for helping me out
anonymous
  • anonymous
no problem anytime
rock_mit182
  • rock_mit182
@IrishBoy123 any thoughts ?
IrishBoy123
  • 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]
rock_mit182
  • rock_mit182
yeah that is what i meant
rock_mit182
  • rock_mit182
distance is halved
IrishBoy123
  • IrishBoy123
so \[4 \times 10^2 = ?? \times 5^2\]

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