anonymous
  • anonymous
If you decrease the distance between two identical point charges so that the new distance is five time the original distance apart, what happens to the force between them?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
A. it is multiplied by 5 B. it is multiplied by 25 C. it is divided by 5 D. it is divided by 25
anonymous
  • anonymous
What is the equation for the force and how is it related to the distance?
anonymous
  • anonymous
i forget :(

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Michele_Laino
  • Michele_Laino
before decreasing the distance we can write: \[\Large F = K\frac{{{Q_1}{Q_2}}}{{{d^2}}}\] whereas after, we can write: \[\Large {F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}}\]
anonymous
  • anonymous
ok! and so what do i plug in? :/
Michele_Laino
  • Michele_Laino
here is next step: \[{F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}} = K\frac{{{Q_1}{Q_2}}}{{25{d^2}}} = \frac{1}{{25}}K\frac{{{Q_1}{Q_2}}}{{{d^2}}} = \frac{1}{{25}}F\]
Michele_Laino
  • Michele_Laino
\[\Large {F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}} = K\frac{{{Q_1}{Q_2}}}{{25{d^2}}} = \frac{1}{{25}}K\frac{{{Q_1}{Q_2}}}{{{d^2}}} = \frac{1}{{25}}F\]
Michele_Laino
  • Michele_Laino
F is the initial force, whereas F_1 is the final force
anonymous
  • anonymous
i cannot see what you wrote on that last part of the equation :/
anonymous
  • anonymous
:(
anonymous
  • anonymous
i see 1/25 and then teh rest disappears :(
anonymous
  • anonymous
@Michele_Laino ?
Michele_Laino
  • Michele_Laino
\[\large {F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}} = K\frac{{{Q_1}{Q_2}}}{{25{d^2}}} = \frac{1}{{25}}K\frac{{{Q_1}{Q_2}}}{{{d^2}}} = \frac{1}{{25}}F\]
Michele_Laino
  • Michele_Laino
the magnitude of the final force is 1/25 of magnitude of the initial force
anonymous
  • anonymous
oh so our solution is D? divided by 25?
Michele_Laino
  • Michele_Laino
yes! that's right!
anonymous
  • anonymous
yay! thank you!!
Michele_Laino
  • Michele_Laino
thank you!! :)

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