Question

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?

Answer 1

A. it is multiplied by 5
B. it is multiplied by 25
C. it is divided by 5
D. it is divided by 25

Answer 2

What is the equation for the force and how is it related to the distance?

Answer 3

i forget :(

Answer 4

before decreasing the distance we can write:
\[\Large F = K\frac{{{Q_1}{Q_2}}}{{{d^2}}}\]
whereas after, we can write:
\[\Large {F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}}\]

Answer 5

ok! and so what do i plug in? :/

Answer 6

here is next step:
\[{F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}} = K\frac{{{Q_1}{Q_2}}}{{25{d^2}}} = \frac{1}{{25}}K\frac{{{Q_1}{Q_2}}}{{{d^2}}} = \frac{1}{{25}}F\]

Answer 7

\[\Large {F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}} = K\frac{{{Q_1}{Q_2}}}{{25{d^2}}} = \frac{1}{{25}}K\frac{{{Q_1}{Q_2}}}{{{d^2}}} = \frac{1}{{25}}F\]

Answer 8

F is the initial force, whereas F_1 is the final force

Answer 9

i cannot see what you wrote on that last part of the equation :/

Answer 10

:(

Answer 11

i see 1/25 and then teh rest disappears :(

Answer 12

@Michele_Laino ?

Answer 13

\[\large {F_1} = K\frac{{{Q_1}{Q_2}}}{{{{\left( {5d} \right)}^2}}} = K\frac{{{Q_1}{Q_2}}}{{25{d^2}}} = \frac{1}{{25}}K\frac{{{Q_1}{Q_2}}}{{{d^2}}} = \frac{1}{{25}}F\]

Answer 14

the magnitude of the final force is 1/25 of magnitude of the initial force

Answer 15

oh so our solution is D? divided by 25?

Answer 16

yes! that's right!

Answer 17

yay! thank you!!

Answer 18

thank you!! :)