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  • DLS

Derive an expression for the G.field due to a uniform rod of Length L and mass M at a point on its perpendicular bisector at a distance d from the centre

Physics
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  • DLS
|dw:1361012885773:dw|
  • DLS
\[\LARGE \frac{2GM}{\sqrt{4L^2+d^2}}\] is what I got one term seems missing
again, seems like you didnt take the components right. E will be given by dE= 2dE1 sintheta where dE1=G(dm)/(d^2 + x^2) where dm=m/l dx sintheta=d/(d^2 + x^2) integrate with respect to x from 0 to l/2 |dw:1361012228614:dw|

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Other answers:

lol.. in such cases its better to take 2 diametrically opposite mass elements!
  • DLS
??
wait wait..lemme think!
@DLS you didnt get what i said?
  • DLS
no ..u wrote scary symbols :P
hmm. haha :P
|dw:1361026132081:dw| something liek this??.. cause if you consider two elements at a time, then the net field due to them becomes perpendicular to the rod.. and hence you can now add up all the elements.. else you couldn't add up directly.. cause, the fields would be in different directions!
|dw:1361026342375:dw|
  • DLS
ek se kyun nahi kar sakte O.O
  • DLS
|dw:1361026384089:dw|
you need to learn more vector analysis.. its a continous mass distribution!..
  • DLS
:/ what
yrelhan please explain.. dls whatever he said .. is exactly correct!
  • DLS
bolo na :/
what should i explain?
ek se kyu nhi kr skte? in these kind of questions we always take 2 elements so that1 component cancel out and we are left with one. easier to integrate. how to do with one element? i dont know.
You can integrate more easily using an angular variable.
Can't you use similar triangles?
I have: \(\LARGE \frac{2GM}{d\sqrt{L^2+4d^2}}\)
  • DLS
i didnt get d :/
  • DLS
i mean whats wrong with my method
  • DLS
The mass is a distance of \[\LARGE \sqrt{L^2+\frac{d^2}{4}}\] from the rod
???? the mass is a distance \(\Large d\) from the rod. What is your method? I have not seen it in your posts.
It's like a problem of electric field. You have to integrate Newton's law of universal gravitation in the same way as you would use Coulomb's law for charges.
  • DLS
so cant we do like this integrate from \[\LARGE \sqrt{L^2+\frac{d^2}{4}}\] to L/2 and multiply it by 2 :O
  • DLS
that would be complicated but eh
  • DLS
why are we considering elemental mass?
This is why I told you to use angle as integration variable.
  • DLS
okay..thanks :O but thats correct right?just complex..thanks!
Wait, I'll send the derivation.
Here it is.
1 Attachment

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