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DLS
 2 years 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
DLS
 2 years 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

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

yrelhan4
 2 years ago
Best ResponseYou've already chosen the best response.0again, 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

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

yrelhan4
 2 years ago
Best ResponseYou've already chosen the best response.0@DLS you didnt get what i said?

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0no ..u wrote scary symbols :P

Mashy
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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!

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0ek se kyun nahi kar sakte O.O

Mashy
 2 years ago
Best ResponseYou've already chosen the best response.0you need to learn more vector analysis.. its a continous mass distribution!..

Mashy
 2 years ago
Best ResponseYou've already chosen the best response.0yrelhan please explain.. dls whatever he said .. is exactly correct!

yrelhan4
 2 years ago
Best ResponseYou've already chosen the best response.0what should i explain?

yrelhan4
 2 years ago
Best ResponseYou've already chosen the best response.0ek 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.

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.1You can integrate more easily using an angular variable.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0Can't you use similar triangles?

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

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0i mean whats wrong with my method

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

VincentLyon.Fr
 2 years ago
Best 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.

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.1It'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
 2 years ago
Best ResponseYou've already chosen the best response.0so 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
 2 years ago
Best ResponseYou've already chosen the best response.0that would be complicated but eh

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0why are we considering elemental mass?

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.1This is why I told you to use angle as integration variable.

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0okay..thanks :O but thats correct right?just complex..thanks!

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.1Wait, I'll send the derivation.
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