anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@Michele_Laino :)
Michele_Laino
  • Michele_Laino
here we have to apply the law of Coulomb: \[\Large F = K\frac{{{q_1}{q_2}}}{{{d^2}}}\]
anonymous
  • anonymous
ok! what do we plug in?

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rvc
  • rvc
do you know for what the symbols stand @iheartfood
anonymous
  • anonymous
yes, i believe so.... i just get confused with inserting :/
rvc
  • rvc
oh tell us what the symbols mean
Michele_Laino
  • Michele_Laino
the next step is: \[\large {q_2} = \frac{{F{d^2}}}{{K{q_1}}} = \frac{{7.2 \times {{\left( {5 \times {{10}^{ - 2}}} \right)}^2}}}{{9 \times {{10}^9} \times 4 \times {{10}^{ - 6}}}} = ...Coulombs\]
anonymous
  • anonymous
we get 5E-7?
anonymous
  • anonymous
is that right? if so, what is the next step? :)
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
what is the sign of q_2?
anonymous
  • anonymous
positive?
Michele_Laino
  • Michele_Laino
yes! since the force is attractive and q_1 is negative
rvc
  • rvc
attractive force exists between opposite charrges
anonymous
  • anonymous
ooh yay! so our solution will be either -2.0 or -4.0 ? how can we find which it is?
Michele_Laino
  • Michele_Laino
our solution is: \[\large 5 \times {10^{ - 7}} = 0.5 \times {10^{ - 6}} = 0.5\mu C\]
rvc
  • rvc
no if one is negative other will be positive
anonymous
  • anonymous
oh ok! yay thank you!! +0.5 it is!!
Michele_Laino
  • Michele_Laino
yes! @rvc
rvc
  • rvc
thanks
Michele_Laino
  • Michele_Laino
that's right! @iheartfood

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