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?

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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
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A. it is multiplied by 5 B. it is multiplied by 25 C. it is divided by 5 D. it is divided by 25
What is the equation for the force and how is it related to the distance?
i forget :(

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

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