InterviewSolution
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InterviewSolution
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Derive an expression fro the gravitational field due to a unifrom rod of length L and M at a point on its perpendicular bisector at a distance d from the centre. |
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Answer» Correct Answer - B::D A small section of rod is condered at x distance mass of the element `=(M/Lxxdx=dm)` `=dE_1=(G(dm)xx1)/((d^2+x^2))=dE_2` Resultant `eE=2dE_1 sintheta` `2xx(G(dm))/(d^2+x^2)xxd/(sqrt(d^2+x^2))` `=(2xxGMxxd dx)/(L(d^2+x^2({(sqrt(d^2+x^2))})` Total gravitatioN/Al field `E=int_0^(L/2) (2Gmd dx)/(L(d^2+x^2)^(3/2))` integrating the above the equation it can be found that `E=(2Gm)/(dsqrt(L^2+4d^2)0` |
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