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anonymous

  • one year ago

An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other charge? (µC = 1.0 × 10^-6 C) -4.0 µC , -2.0 µC , +2.0 µC , or +4.0 µC

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  1. anonymous
    • one year ago
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    i think it's C.

  2. Michele_Laino
    • one year ago
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    here we have to apply teh law of Coulomb

  3. Michele_Laino
    • one year ago
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    the*

  4. anonymous
    • one year ago
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    Not so sure! sorry!

  5. Michele_Laino
    • one year ago
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    \[\Large F = K\frac{{{Q_1}{Q_2}}}{{{d^2}}}\]

  6. anonymous
    • one year ago
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    +2.0 µC

  7. anonymous
    • one year ago
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    okay! what do we plug in?

  8. Michele_Laino
    • one year ago
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    for example I will solve that fromula for Q2, and I get: \[{Q_2} = \frac{{F{d^2}}}{{K{Q_1}}}\]

  9. anonymous
    • one year ago
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    7.2 should be plugged in

  10. anonymous
    • one year ago
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    ohh what would that look like? :/ i am confused :/ is is 7.2 divided by somehing?

  11. Michele_Laino
    • one year ago
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    substituting our data, we can write: \[\Large {Q_2} = \frac{{F{d^2}}}{{K{Q_1}}} = \frac{{7.2 \times {{\left( {{{10}^{ - 1}}} \right)}^2}}}{{9 \times {{10}^9} \times \left( { 4 \times {{10}^{ - 6}}} \right)}}\]

  12. anonymous
    • one year ago
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    F = k*q1*q2/r^2 so q2 = F*r^2/(k*q1) = 7.2*0.10^2/(9.0x10^9*4.0x10^-6) = 2.0x10^-6 and since the force is attractive the charge must be C - 2.0 µC

  13. anonymous
    • one year ago
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    sorry b! not c!

  14. Michele_Laino
    • one year ago
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    no, it is C, since the force is attractive, and only opposite charges interact with an attractive force

  15. anonymous
    • one year ago
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    ohh so our solution is choice C?

  16. Michele_Laino
    • one year ago
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    yes!

  17. anonymous
    • one year ago
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    yay! thank you both!!

  18. anonymous
    • one year ago
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    your welcome!

  19. anonymous
    • one year ago
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    was that your last question?

  20. anonymous
    • one year ago
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    oh never mind

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