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## anonymous one year ago An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other charge? (µC = 1.0 × 10^-6 C) -4.0 µC , -2.0 µC , +2.0 µC , or +4.0 µC

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1. anonymous

i think it's C.

2. Michele_Laino

here we have to apply teh law of Coulomb

3. Michele_Laino

the*

4. anonymous

Not so sure! sorry!

5. Michele_Laino

$\Large F = K\frac{{{Q_1}{Q_2}}}{{{d^2}}}$

6. anonymous

+2.0 µC

7. anonymous

okay! what do we plug in?

8. Michele_Laino

for example I will solve that fromula for Q2, and I get: ${Q_2} = \frac{{F{d^2}}}{{K{Q_1}}}$

9. anonymous

7.2 should be plugged in

10. anonymous

ohh what would that look like? :/ i am confused :/ is is 7.2 divided by somehing?

11. Michele_Laino

substituting our data, we can write: $\Large {Q_2} = \frac{{F{d^2}}}{{K{Q_1}}} = \frac{{7.2 \times {{\left( {{{10}^{ - 1}}} \right)}^2}}}{{9 \times {{10}^9} \times \left( { 4 \times {{10}^{ - 6}}} \right)}}$

12. anonymous

F = k*q1*q2/r^2 so q2 = F*r^2/(k*q1) = 7.2*0.10^2/(9.0x10^9*4.0x10^-6) = 2.0x10^-6 and since the force is attractive the charge must be C - 2.0 µC

13. anonymous

sorry b! not c!

14. Michele_Laino

no, it is C, since the force is attractive, and only opposite charges interact with an attractive force

15. anonymous

ohh so our solution is choice C?

16. Michele_Laino

yes!

17. anonymous

yay! thank you both!!

18. anonymous

your welcome!

19. anonymous

was that your last question?

20. anonymous

oh never mind

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