## anonymous one year ago Please help I will medal! I am confused on how to solve this question: A thundercloud has an electric charge of 43.2 C near the top of the cloud and -38.7 near the bottom of the cloud. The magnitude of the electric force between these two charges is 3.95 x 10^6 N. What is the average separation between these charges? (kc=8.99 x 10^9 N m^2/C^2)

1. anonymous

@iGreen @gleem @emma.monsterr

2. anonymous

Im sorry, i didnt learn these yet but good luck :/

3. anonymous

ok thank you :/

4. anonymous

@OpenStudyRocks5* @Greg_D @horsegirl325

5. anonymous

The only thing I can think of is to subtract maybe but that comes out too 81.5 and I dont think thats right

6. anonymous

Hi! You can use the Coulomb's force law here...

7. IrishBoy123

seems they want you to use coulumbs law F = k q1 q2 / r^2

8. anonymous

I believe the equation I am supposed to use is in the question above. However, I dont understand how to solve with this equation.

9. anonymous

@Greg_D

10. anonymous

note that you are given both charges: $q_1=43.2C$ $q_2=-38.7C$ and $F=3.95\times 10^6 N$ you are also given kc, which is k in the eq of IrishBoy you can just replace and finde the value for r, which represents the distance.... give it a try!

11. anonymous

So I would solve 3.95 x 10^6 N =8.99 x 10^9 N m^2/C^2 (43.2C)(-38.7C) / r^2?

12. anonymous

yeah!! get r from there

13. anonymous

Ok can you help me solve it? Just so I know I dont get the wrong answer?

14. anonymous

we can both make calculations and check togheter... let me see what i get...

15. anonymous

Alright I'll let you know what I get.

16. anonymous

just a detail, we dont need to use the minus sign for the second charge!

17. anonymous

Alright!

18. anonymous

I got 1.95 x 10^3 what did you get?

19. anonymous

the same!!! :) what do you thinks the units should be?

20. anonymous

m?

21. anonymous

sure! it is a distence, so meters!

22. anonymous

good work :)

23. anonymous

Thank you!

24. anonymous

This correct right? @glittergurl0101

25. anonymous

Yes, this is correct :) @Ella31224