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

- anonymous

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

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- anonymous

|dw:1327758982728:dw|

- anonymous

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

- anonymous

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

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## More answers

- anonymous

450*10^-9 N

- anonymous

can you explain?

- anonymous

it's not the right answer

- anonymous

why?

- Shayaan_Mustafa

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

- Shayaan_Mustafa

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

- anonymous

sure and can you please show steps??

- Shayaan_Mustafa

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

- anonymous

ok thank you

- Shayaan_Mustafa

you are welcome.

- Shayaan_Mustafa

here is diagram first look at this and understand this.

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- Shayaan_Mustafa

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.

- anonymous

looks good

- anonymous

thank you Shayan

- Shayaan_Mustafa

wait i have to go and just come in 2min

- Shayaan_Mustafa

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.

- Shayaan_Mustafa

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.

- Shayaan_Mustafa

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

- Shayaan_Mustafa

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

- Shayaan_Mustafa

finally. sum forces and you will get your answer.

- Shayaan_Mustafa

have you got it? it is done.

- anonymous

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

- Shayaan_Mustafa

can you show your solution?

- anonymous

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

- anonymous

my answer was 4.07*10^-3

- Shayaan_Mustafa

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.

- Shayaan_Mustafa

any question?

- anonymous

shayan can you help me?

- Shayaan_Mustafa

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

- anonymous

thank you i'll try again!

- Shayaan_Mustafa

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

- anonymous

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

- Shayaan_Mustafa

ok w8. let me analyze.

- Shayaan_Mustafa

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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- Shayaan_Mustafa

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

- anonymous

zor zor supas aziyatm dait

- Shayaan_Mustafa

kia?

- Shayaan_Mustafa

what is this language?

- anonymous

what is your language?

- Shayaan_Mustafa

urdu and english.

- anonymous

I am kurdish

- Shayaan_Mustafa

ok.
now good luck. take care everyone.

- anonymous

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

- Shayaan_Mustafa

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

- anonymous

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

- anonymous

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.

- anonymous

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

- Shayaan_Mustafa

elica85 are you here?

- Shayaan_Mustafa

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.

- Shayaan_Mustafa

thanks to Jemurray3.

- anonymous

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

- Shayaan_Mustafa

hi jemurray3

- anonymous

Hi there Shayaan

- Shayaan_Mustafa

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

- anonymous

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 ^^

- Shayaan_Mustafa

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

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