## 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?

1. 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

2. anonymous

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

3. anonymous

i forget :(

4. Michele_Laino

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

5. anonymous

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

6. Michele_Laino

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

7. Michele_Laino

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

8. Michele_Laino

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

9. anonymous

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

10. anonymous

:(

11. anonymous

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

12. anonymous

@Michele_Laino ?

13. Michele_Laino

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

14. Michele_Laino

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

15. anonymous

oh so our solution is D? divided by 25?

16. Michele_Laino

yes! that's right!

17. anonymous

yay! thank you!!

18. Michele_Laino

thank you!! :)