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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 }

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  1. anonymous
    • one year ago
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    i can help you

  2. rock_mit182
    • one year ago
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    \[4 = k \frac{ q*q }{r^2 } \]

  3. rock_mit182
    • one year ago
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    0,1 meters for the initial values

  4. rock_mit182
    • one year ago
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    now if the radius is multiplied by 1/2

  5. anonymous
    • one year ago
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    yes

  6. rock_mit182
    • one year ago
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    the radius would be 1/20, right ?

  7. anonymous
    • one year ago
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    yea you got it keep going

  8. rock_mit182
    • one year ago
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    im stuck

  9. rock_mit182
    • one year ago
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    i think the right options are 8N or 16 N

  10. anonymous
    • one year ago
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    16 N

  11. rock_mit182
    • one year ago
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    because r is squared, right ?

  12. anonymous
    • one year ago
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    yea

  13. rock_mit182
    • one year ago
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    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
    • one year ago
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    \[4 = k \frac{ 2q*2q }{ (0,1)^2 }\]

  15. anonymous
    • one year ago
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    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
    • one year ago
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    you get it?

  17. rock_mit182
    • one year ago
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    what did you do ?

  18. rock_mit182
    • one year ago
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    did you see my last equation ?

  19. rock_mit182
    • one year ago
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    i curious about what just did happen

  20. anonymous
    • one year ago
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    which one?

  21. rock_mit182
    • one year ago
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    \[4 =k \frac{ 2q *2q }{ (0,1)^2 }\]

  22. rock_mit182
    • one year ago
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    F1 =2 so F2 = ? ; keeping in mind that Im using the initial values but double charge ( each one)

  23. rock_mit182
    • one year ago
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    im sorry F1 =4 N

  24. anonymous
    • one year ago
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    oka what is the question asking? id dont see

  25. rock_mit182
    • one year ago
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    F2 is incresing respect to one by a factor of ?

  26. anonymous
    • one year ago
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    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
    • one year ago
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    and i think is by a factor of 4

  28. rock_mit182
    • one year ago
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    anyway thanks a lot for helping me out

  29. anonymous
    • one year ago
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    no problem anytime

  30. rock_mit182
    • one year ago
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    @IrishBoy123 any thoughts ?

  31. IrishBoy123
    • one year ago
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    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
    • one year ago
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    yeah that is what i meant

  33. rock_mit182
    • one year ago
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    distance is halved

  34. IrishBoy123
    • one year ago
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    so \[4 \times 10^2 = ?? \times 5^2\]

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