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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
 one year ago
 one year ago
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
 one year ago
 one year ago

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DLSBest ResponseYou've already chosen the best response.0
\[\LARGE \frac{2GM}{\sqrt{4L^2+d^2}}\] is what I got one term seems missing
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
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
 one year ago

MashyBest ResponseYou've already chosen the best response.0
lol.. in such cases its better to take 2 diametrically opposite mass elements!
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
@DLS you didnt get what i said?
 one year ago

DLSBest ResponseYou've already chosen the best response.0
no ..u wrote scary symbols :P
 one year ago

MashyBest ResponseYou've already chosen the best response.0
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!
 one year ago

DLSBest ResponseYou've already chosen the best response.0
ek se kyun nahi kar sakte O.O
 one year ago

MashyBest ResponseYou've already chosen the best response.0
you need to learn more vector analysis.. its a continous mass distribution!..
 one year ago

MashyBest ResponseYou've already chosen the best response.0
yrelhan please explain.. dls whatever he said .. is exactly correct!
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
what should i explain?
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
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.
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
You can integrate more easily using an angular variable.
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
Can't you use similar triangles?
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
I have: \(\LARGE \frac{2GM}{d\sqrt{L^2+4d^2}}\)
 one year ago

DLSBest ResponseYou've already chosen the best response.0
i mean whats wrong with my method
 one year ago

DLSBest ResponseYou've already chosen the best response.0
The mass is a distance of \[\LARGE \sqrt{L^2+\frac{d^2}{4}}\] from the rod
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
???? the mass is a distance \(\Large d\) from the rod. What is your method? I have not seen it in your posts.
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
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.
 one year ago

DLSBest ResponseYou've already chosen the best response.0
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
 one year ago

DLSBest ResponseYou've already chosen the best response.0
that would be complicated but eh
 one year ago

DLSBest ResponseYou've already chosen the best response.0
why are we considering elemental mass?
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
This is why I told you to use angle as integration variable.
 one year ago

DLSBest ResponseYou've already chosen the best response.0
okay..thanks :O but thats correct right?just complex..thanks!
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
Wait, I'll send the derivation.
 one year ago
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