## elica85 Group Title What is the magnitude of the force on the -10 charge in the figure ? 2 years ago 2 years ago

1. elica85 Group Title

|dw:1327758982728:dw|

2. elica85 Group Title

the resultant force should be...|dw:1327759214456:dw| something like that

3. elica85 Group Title

broke down x and y components for force on + on - and - on - and then added...not getting right answer

450*10^-9 N

5. elica85 Group Title

can you explain?

6. elica85 Group Title

why?

8. Shayaan_Mustafa Group Title

it is not easy. it is a vector quantity. therefore you must have to consider directions.

9. Shayaan_Mustafa Group Title

you have to wait until i solve it for you. ok elicia85?

10. elica85 Group Title

sure and can you please show steps??

11. Shayaan_Mustafa Group Title

yes i will show steps to you. wait. i am here.

12. elica85 Group Title

ok thank you

13. Shayaan_Mustafa Group Title

you are welcome.

14. Shayaan_Mustafa Group Title

here is diagram first look at this and understand this.

15. Shayaan_Mustafa Group Title

is this pic making sense to you? if i do anything wrong here then kindly point out me. you too faryad, hei, elicia85 and an unknown person.

16. elica85 Group Title

looks good

thank you Shayan

18. Shayaan_Mustafa Group Title

wait i have to go and just come in 2min

19. Shayaan_Mustafa Group Title

hello. it is taking too much time for me. lets solve step by step here. ok. force of +15nC will be represented by F15 force of -10nC will be represented by F10 force of -5nC will be represented by F5 ok.

20. Shayaan_Mustafa Group Title

now as you can see. on -10nC there are two forces acting on it. 1)F15 which is attractive and at an angle of 45 2)F5 which repulsive.

21. Shayaan_Mustafa Group Title

F15 can be resolve into factors. along x-axis and along y-axis.

22. Shayaan_Mustafa Group Title

forces along y-axis can be calculated as, F15y=(kq1q2)/r² sin45 and F5y=(kq1q2)/r² forces along x-axis can be calculated as, F15x=(kq1q2)/r² cos45

23. Shayaan_Mustafa Group Title

24. Shayaan_Mustafa Group Title

have you got it? it is done.

25. elica85 Group Title

it's exactly what i did but i will try again

26. Shayaan_Mustafa Group Title

27. elica85 Group Title

well you still have to add the F5 on F10 don't you? which will only consist of y direction force

28. elica85 Group Title

29. Shayaan_Mustafa Group Title

well you still have to add the F5 on F10 don't you? which will only consist of y direction force on y-axis: F10y=F15sin45 - F5 on x-axis: F10x= -F15x finally, $F _{10}=\sqrt{(F _{10y})^{2}+(F _{10x})^{2}}$ This is complete.

30. Shayaan_Mustafa Group Title

any question?

shayan can you help me?

32. Shayaan_Mustafa Group Title

of course fryad. although we are brothers. arn't we?

33. elica85 Group Title

thank you i'll try again!

34. Shayaan_Mustafa Group Title

you are ever welcome. Regards, Read my Profile. :D

yes ,sure thank you , please prove this for me A.B=cos& |A|.|B|

36. Shayaan_Mustafa Group Title

ok w8. let me analyze.

37. Shayaan_Mustafa Group Title

The three vectors above form the triangle AOB. using Law of Cosines , $|a-b|^{2}=|a|^{2}+|b|^{2}-2|a||b|\cos \theta$ using properties of dot product. $|a-b|^{2}=(a-b).(a-b)$ $=a.a-a.b-b.a+b.b$ $=|a|^{2}+|b|^{2}-2a.b$ original becomes, $|a-b|^{2}=|a|^{2}+|b ^{2}|-2|a||b|\cos \theta$ and, $|a|^{2}-2a.b+|b|^{2}=|a|^{2}+|b ^{2}|-2|a||b|\cos \theta$ now cancel terms which are same and you will got $a.b=|a||b|\cos \theta$

38. Shayaan_Mustafa Group Title

that is all. Regards, Read my Profile. :D

zor zor supas aziyatm dait

40. Shayaan_Mustafa Group Title

kia?

41. Shayaan_Mustafa Group Title

what is this language?

43. Shayaan_Mustafa Group Title

urdu and english.

I am kurdish

45. Shayaan_Mustafa Group Title

ok. now good luck. take care everyone.

I am beginner in physic,thus I want help me for any time if you can my brother?

47. Shayaan_Mustafa Group Title

sure sure. i am often available here. you can post your question.

That is all kind of you for every thing.can I recognize you in facebook?

49. Jemurray3 Group Title

Just skimming through this.... You should recognize that the angle is not 45 degrees because it's a rectangle, not a square, so the values above would not yield the correct answer.

50. elica85 Group Title

good point, i wasn't using 45 though, just used the measurements given. i still need to resolve, haven't got to it yet

51. Shayaan_Mustafa Group Title

elica85 are you here?

52. Shayaan_Mustafa Group Title

oh yes. so angle is not 45degree. it can be calculated as, $\tan^{-1} (0.01/0.03)=18.43degrees.$ Kindly use this angle instead of 45.

53. Shayaan_Mustafa Group Title

thanks to Jemurray3.

54. Jemurray3 Group Title

elica I assume you are adding the components separately and then using the Pythagorean Theorem to calculate the magnitude of the resultant, right?

55. Shayaan_Mustafa Group Title

hi jemurray3

56. Jemurray3 Group Title

Hi there Shayaan

57. Shayaan_Mustafa Group Title

http://openstudy.com/study#/updates/4f245a39e4b0a2a9c2667f2f We need you in mathematics section. kindly help us.

58. elica85 Group Title

thank you everyone, i did get the right answer finally. i was confused about +/-. i thought the - charge on negative would make it negative direction going down y, making it double negative...so i was just getting calculations wrong. it was 18 degrees ^^

59. Shayaan_Mustafa Group Title

you are ever welcome. we are made for help. pleasure.