## DLS Group Title 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

1. DLS Group Title

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2. DLS Group Title

$\LARGE \frac{2GM}{\sqrt{4L^2+d^2}}$ is what I got one term seems missing

3. yrelhan4 Group Title

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|

4. Mashy Group Title

lol.. in such cases its better to take 2 diametrically opposite mass elements!

5. DLS Group Title

??

6. Mashy Group Title

wait wait..lemme think!

7. yrelhan4 Group Title

@DLS you didnt get what i said?

8. DLS Group Title

no ..u wrote scary symbols :P

9. yrelhan4 Group Title

hmm. haha :P

10. Mashy Group Title

|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!

11. Mashy Group Title

|dw:1361026342375:dw|

12. DLS Group Title

ek se kyun nahi kar sakte O.O

13. DLS Group Title

|dw:1361026384089:dw|

14. Mashy Group Title

you need to learn more vector analysis.. its a continous mass distribution!..

15. DLS Group Title

:/ what

16. Mashy Group Title

yrelhan please explain.. dls whatever he said .. is exactly correct!

17. DLS Group Title

bolo na :/

18. yrelhan4 Group Title

what should i explain?

19. yrelhan4 Group Title

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.

20. Vincent-Lyon.Fr Group Title

You can integrate more easily using an angular variable.

21. abb0t Group Title

Can't you use similar triangles?

22. Vincent-Lyon.Fr Group Title

I have: $$\LARGE \frac{2GM}{d\sqrt{L^2+4d^2}}$$

23. DLS Group Title

i didnt get d :/

24. DLS Group Title

i mean whats wrong with my method

25. DLS Group Title

The mass is a distance of $\LARGE \sqrt{L^2+\frac{d^2}{4}}$ from the rod

26. Vincent-Lyon.Fr Group Title

???? the mass is a distance $$\Large d$$ from the rod. What is your method? I have not seen it in your posts.

27. Vincent-Lyon.Fr Group Title

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.

28. DLS Group Title

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

29. DLS Group Title

that would be complicated but eh

30. DLS Group Title

why are we considering elemental mass?

31. Vincent-Lyon.Fr Group Title

This is why I told you to use angle as integration variable.

32. DLS Group Title

okay..thanks :O but thats correct right?just complex..thanks!

33. Vincent-Lyon.Fr Group Title

Wait, I'll send the derivation.

34. Vincent-Lyon.Fr Group Title

Here it is.