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anonymous
 one year ago
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?
anonymous
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A. it is multiplied by 5 B. it is multiplied by 25 C. it is divided by 5 D. it is divided by 25

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the equation for the force and how is it related to the distance?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1before 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
 one year ago
Best ResponseYou've already chosen the best response.0ok! and so what do i plug in? :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large {F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}} = K\frac{{{Q_1}{Q_2}}}{{25{d^2}}} = \frac{1}{{25}}K\frac{{{Q_1}{Q_2}}}{{{d^2}}} = \frac{1}{{25}}F\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1F is the initial force, whereas F_1 is the final force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i cannot see what you wrote on that last part of the equation :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i see 1/25 and then teh rest disappears :(

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\large {F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}} = K\frac{{{Q_1}{Q_2}}}{{25{d^2}}} = \frac{1}{{25}}K\frac{{{Q_1}{Q_2}}}{{{d^2}}} = \frac{1}{{25}}F\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the magnitude of the final force is 1/25 of magnitude of the initial force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh so our solution is D? divided by 25?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! that's right!
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