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anonymous

  • one year ago

An attractive force of 7.2 N occurs between two point charges that are 0.05 m apart. If one charge is - 4.0 µC, what is the other charge? A. +0.5 µC B. +2.0 µC C. -2.0 µC D. -4.0 µC

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  1. anonymous
    • one year ago
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    @Michele_Laino :)

  2. Michele_Laino
    • one year ago
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    here we have to apply the law of Coulomb: \[\Large F = K\frac{{{q_1}{q_2}}}{{{d^2}}}\]

  3. anonymous
    • one year ago
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    ok! what do we plug in?

  4. rvc
    • one year ago
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    do you know for what the symbols stand @iheartfood

  5. anonymous
    • one year ago
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    yes, i believe so.... i just get confused with inserting :/

  6. rvc
    • one year ago
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    oh tell us what the symbols mean

  7. Michele_Laino
    • one year ago
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    the next step is: \[\large {q_2} = \frac{{F{d^2}}}{{K{q_1}}} = \frac{{7.2 \times {{\left( {5 \times {{10}^{ - 2}}} \right)}^2}}}{{9 \times {{10}^9} \times 4 \times {{10}^{ - 6}}}} = ...Coulombs\]

  8. anonymous
    • one year ago
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    we get 5E-7?

  9. anonymous
    • one year ago
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    is that right? if so, what is the next step? :)

  10. Michele_Laino
    • one year ago
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    that's right!

  11. Michele_Laino
    • one year ago
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    what is the sign of q_2?

  12. anonymous
    • one year ago
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    positive?

  13. Michele_Laino
    • one year ago
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    yes! since the force is attractive and q_1 is negative

  14. rvc
    • one year ago
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    attractive force exists between opposite charrges

  15. anonymous
    • one year ago
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    ooh yay! so our solution will be either -2.0 or -4.0 ? how can we find which it is?

  16. Michele_Laino
    • one year ago
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    our solution is: \[\large 5 \times {10^{ - 7}} = 0.5 \times {10^{ - 6}} = 0.5\mu C\]

  17. rvc
    • one year ago
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    no if one is negative other will be positive

  18. anonymous
    • one year ago
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    oh ok! yay thank you!! +0.5 it is!!

  19. Michele_Laino
    • one year ago
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    yes! @rvc

  20. rvc
    • one year ago
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    thanks

  21. Michele_Laino
    • one year ago
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    that's right! @iheartfood

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