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What is the magnitude of the force on the -10 charge in the figure ?

Physics
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|dw:1327758982728:dw|
the resultant force should be...|dw:1327759214456:dw| something like that
broke down x and y components for force on + on - and - on - and then added...not getting right answer

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Other answers:

450*10^-9 N
can you explain?
it's not the right answer
why?
it is not easy. it is a vector quantity. therefore you must have to consider directions.
you have to wait until i solve it for you. ok elicia85?
sure and can you please show steps??
yes i will show steps to you. wait. i am here.
ok thank you
you are welcome.
here is diagram first look at this and understand this.
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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.
looks good
thank you Shayan
wait i have to go and just come in 2min
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.
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.
F15 can be resolve into factors. along x-axis and along y-axis.
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
finally. sum forces and you will get your answer.
have you got it? it is done.
it's exactly what i did but i will try again
can you show your solution?
well you still have to add the F5 on F10 don't you? which will only consist of y direction force
my answer was 4.07*10^-3
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.
any question?
shayan can you help me?
of course fryad. although we are brothers. arn't we?
thank you i'll try again!
you are ever welcome. Regards, Read my Profile. :D
yes ,sure thank you , please prove this for me A.B=cos& |A|.|B|
ok w8. let me analyze.
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 \[a.b=|a||b|\cos \theta\]
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that is all. Regards, Read my Profile. :D
zor zor supas aziyatm dait
kia?
what is this language?
what is your language?
urdu and english.
I am kurdish
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?
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?
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.
good point, i wasn't using 45 though, just used the measurements given. i still need to resolve, haven't got to it yet
elica85 are you here?
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.
thanks to Jemurray3.
elica I assume you are adding the components separately and then using the Pythagorean Theorem to calculate the magnitude of the resultant, right?
hi jemurray3
Hi there Shayaan
http://openstudy.com/study#/updates/4f245a39e4b0a2a9c2667f2f We need you in mathematics section. kindly help us.
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 ^^
you are ever welcome. we are made for help. pleasure.

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