## anonymous one year ago An attractive force of 7.2 N occurs between two point charges that are 0.05 m apart. If one charge is - 4.0 µC, what is the other charge? A. +0.5 µC B. +2.0 µC C. -2.0 µC D. -4.0 µC

1. anonymous

@Michele_Laino :)

2. Michele_Laino

here we have to apply the law of Coulomb: $\Large F = K\frac{{{q_1}{q_2}}}{{{d^2}}}$

3. anonymous

ok! what do we plug in?

4. rvc

do you know for what the symbols stand @iheartfood

5. anonymous

yes, i believe so.... i just get confused with inserting :/

6. rvc

oh tell us what the symbols mean

7. Michele_Laino

the next step is: $\large {q_2} = \frac{{F{d^2}}}{{K{q_1}}} = \frac{{7.2 \times {{\left( {5 \times {{10}^{ - 2}}} \right)}^2}}}{{9 \times {{10}^9} \times 4 \times {{10}^{ - 6}}}} = ...Coulombs$

8. anonymous

we get 5E-7?

9. anonymous

is that right? if so, what is the next step? :)

10. Michele_Laino

that's right!

11. Michele_Laino

what is the sign of q_2?

12. anonymous

positive?

13. Michele_Laino

yes! since the force is attractive and q_1 is negative

14. rvc

attractive force exists between opposite charrges

15. anonymous

ooh yay! so our solution will be either -2.0 or -4.0 ? how can we find which it is?

16. Michele_Laino

our solution is: $\large 5 \times {10^{ - 7}} = 0.5 \times {10^{ - 6}} = 0.5\mu C$

17. rvc

no if one is negative other will be positive

18. anonymous

oh ok! yay thank you!! +0.5 it is!!

19. Michele_Laino

yes! @rvc

20. rvc

thanks

21. Michele_Laino

that's right! @iheartfood